What is the sum of 1/(2+ √5) +1/(√5+ √6).......+1/(√99+ √100) ?

An Inequality: $\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\ldots\cdot\frac{99}{100} < \frac{1}{10}$

A product of fractions $\displaystyle \frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\ldots\cdot\frac{2n-1}{2n}$ is on the left-hand side of several inequalities: one with a beautiful proof, one that strengthens the former but is virtually impossible to prove, and a third, even stronger, with an elementary proof.

Try your hand with the simplest variation:

[1]

$\displaystyle \frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\ldots\cdot\frac{99}{100} \lt\frac{1}{10}.$

Solution

|Contact| |Front page| |Contents| |Up|

Copyright © 1996-2018 Alexander Bogomolny

Denote the left-hand side of the inequality A:

$\displaystyle A = \frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\ldots \cdot\frac{99}{100}.$

And introduce its nemesis $B$:

$\displaystyle B = \frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot\ldots\cdot\frac{98}{99}.$

Factor by factor, the fractions in $B$ exceed those in $A:$

$\displaystyle \frac{2}{3} \gt \frac{1}{2},$ $\displaystyle \frac{4}{5} \gt \frac{3}{4},\ldots,\frac{98}{99} \gt \frac{97}{98},$ $\displaystyle 1 \gt \frac{99}{100}.$

From this it follows that $A \lt B.$ Note that, due to the choice of $B,$ in the product $AB$ most of the terms cancel out: $\displaystyle AB = \frac{1}{100}.$ From here,

$\displaystyle A^{2} \lt AB = \frac{1}{100},$

which, with one additional step, proves [1].

This proof suggests that [1] is in fact just a special case of a more general inequality

[2]

$\displaystyle \frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\ldots \cdot\frac{2n-1}{2n} \lt \frac{1}{\sqrt{2n}},$

whose proof is a slight modification of the above with $A$ and $B$ defined as

$\displaystyle A[n] = \frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\ldots \cdot\frac{2n-1}{2n},\\ \displaystyle B[n] = \frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot\ldots \cdot\frac{2n-2}{2n-1}.$

As we shall see shortly, [1] and [2] are quite weak: $A[n]$ has a much better bound, viz.

[3]

$\displaystyle A[n] \lt\frac{1}{\sqrt{3n+1}}.$

[3] supplies an edifying curiosity. By itself, it is easily proved by mathematical induction. However, its weakened version

[3']

$\displaystyle A[n] \lt\frac{1}{\sqrt{3n}},$

as far as I know, does not submit to an inductive proof. Try it, by all means. [3] and [3'] are often quoted as a pair of problems of which the harder one has a simpler proof.

Meanwhile here's a proof for [3].

|Contact| |Front page| |Contents| |Up|

Copyright © 1996-2018 Alexander Bogomolny

To remind,

$\displaystyle A[n] = \frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\ldots \cdot\frac{2n-1}{2n}$

and we wish to prove [3]: $\displaystyle A[n] \lt\frac{1}{\sqrt{3n+1}}.$ For $n = 1,$ we have

$\displaystyle A[1] = \frac{1}{2} = \frac{1}{\sqrt{3\cdot 1+1}}.$

But already for $n = 2,$

$\displaystyle A[2] = \frac{1}{2}\cdot\frac{3}{4} = \frac{3}{8} \lt\frac{1}{\sqrt{7}} = \frac{1}{\sqrt{3\cdot 2+1}},$

because upon squaring $\displaystyle \frac{9}{64} \lt\frac{1}{7},$ for $7\cdot 9 = 63 \lt 64.$ Thus let's proceed with the inductive step and assume that [3] holds for $n = k:$

[4]

$\displaystyle A[k] \lt\frac{1}{\sqrt{3k+1}}.$

We are going to prove that, for $n = k+1,$ [3] also holds

[5]

$\displaystyle A[k+1] \lt\frac{1}{\sqrt{3[k+1]+1}} =\frac{1}{\sqrt{3k+4}}.$

Since $\displaystyle A[k+1] = A[k]\cdot\frac{2k+1}{2k+2},$ [4] implies

[6]

$\displaystyle A[k+1] \lt\frac{2k+1}{2k+2}\cdot\frac{1}{\sqrt{3k+1}}.$

Now square the right hand side in [6]:

$\displaystyle \begin{align} \left[\frac{2k+1}{2k+2}\cdot\frac{1}{\sqrt{3k+1}}\right]^{2}&= \frac{[2k+1]^{2}}{[2k+2]^{2}[3k+1]}\\ &= \frac{[2k+1]^{2}}{12k^{3} + 28k^{2} + 20k + 4}\\ &= \frac{[2k+1]^{2}}{[12k^{3} + 28k^{2} + 19k + 4] + k}\\ &= \frac{[2k+1]^{2}}{[2k+1]^{2}[3k+4] + k}\\ &\lt\frac{[2k+1]^{2}}{[2k+1]^{2}[3k+4]}\\ &= \frac{1}{3k+4}, \end{align}$

which is exactly the right-hand side of [5] and proves [6].

Curiously, a much weaker $\displaystyle A[n] \lt\frac{1}{\sqrt{n}}$ is still resistant to the inductive argument, whereas a stronger version $\displaystyle A[n] \lt\frac{1}{\sqrt{n + 1}}$ goes through without a hitch.

[There is another example where mathematical induction applies easily to a stronger inequality and does not seem to work for a weaker one.]

References

1. A. Engel, Problem-Solving Strategies, Springer Verlag, 1998, p. 180
2. D. Fomin,S. Genkin,I. Itenberg, Mathematical Circles [Russian Experience], AMS, 1996, p. 90
3. S. Savchev, T. Andreescu, Mathematical Miniatures, MAA, 2003, p. 51
4. D. O. Shklyarsky, N. N. Chentsov, I. M. Yaglom, Selected Problems and Theorems of Elementary Mathematics, v 1, Moscow, 1959. [In Russian]

• Schur's Inequality $[x^t[x-y][x-z]+y^t[y-z][y-x]+z^t[z-x][z-y]\ge 0]$
  • An Application of Schur's Inequality $\small{\left[5\left[\sum\sqrt{x+y}\right]\left[\sum\sqrt{[x+y][y+z]}\right]\ge\left[\sum\sqrt{x+y}\right]^3+18\sqrt{[x+y][y+z][z+x]}\right]}$
  • Powers and Fractions Inequality $\left[\displaystyle\sum_{cycl}\frac{a^3b^3}{c^5}\ge \sum_{cycl}\frac{ab}{c}\right]$
  • Three Junior Problems from Vietnam $\left[\left[\begin{array}{ccc}0&y&z\\x&0&z\\x&y&0\end{array}\right]\left[\begin{array}\;a\\b\\c\end{array}\right]=\left[\begin{array}\;x\\y\\z\end{array}\right]\right]$
• Newton's and Maclaurin's Inequalities $\left[E_k[X]\right]^2\gt E_{k-1}[X]E_{k+1}[X],$ $\left[\left[E_k[X]\right]^2\gt E_{k-1}[X]E_{k+1}[X]\right]$
• Rearrangement Inequality $\left[\displaystyle\sum_{i=1}^{n}x_iy_{n+1-i}\le\sum_{i=1}^{n}x_iy_{\sigma[i]}\le\sum_{i=1}^{n}x_iy_i\right]$
• Chebyshev Inequality $\left[\displaystyle n\sum_{i=1}^{n}x_iy_{i}\ge\sum_{i=1}^{n}x_i\sum_{i=1}^{n}y_i\right]$
  • Chebyshev Holds a Key $\left[\displaystyle\sum_{k=1}^n\left[\left[\sum_{i=1}^k\sin a_i\right]\left[\sum_{i=1}^k\cos a_i\right]\right]\le\frac{n[n+1][2n+1]}{12}\right]$
• Jensen's Inequality $\left[ f[\lambda x_1+[1-\lambda ]x_2]\le \lambda f[x_1]+[1-\lambda ]f[x_2]\right]$
• Muirhead's Inequality $\left[\mathbf{a}\succ\mathbf{b}\;\right]$ implies $\left[[\mathbf{a}]\ge [\mathbf{b}]\right]$
• Bergström's inequality $\left[\displaystyle \frac{x_1^2}{a_1}+\frac{x_2^2}{a_2}+\cdots+\frac{x_n^2}{a_n}\ge\frac{[x_1+x_2+\cdots+x_n]^2}{a_1+a_2+\cdots+a_n}\right]$
  • Dorin Marghidanu's Example II $\left[\displaystyle \sum_{k=1}^n\frac{a^2_k}{a_{\sigma[k]}-1}\ge 4n\right]$
• Radon's Inequality and Applications $\left[\displaystyle \frac{x_1^{p+1}}{a_1^p}+\frac{x_2^{p+1}}{a_2^p}+\cdots+\frac{x_n^{p+1}}{a_n^p}\ge\frac{[x_1+x_2+\cdots+x_n]^{p+1}}{[a_1+a_2+\cdots+a_n]^p}\right]$
  • Dorin Marghidanu's Example for Radon's Inequality $\left[\displaystyle\frac{a^{m+n+1}}{b^mc^n}+\frac{b^{m+n+1}}{c^ma^n}+\frac{c^{m+n+1}}{a^mb^n}\ge a+b+c\right]$
• Jordan and Kober Inequalities, PWW $\left[\displaystyle x\in\left[0,\frac{\pi}{2}\right]\,\Longrightarrow\,\sin x\gt\frac{2x}{\pi}\right]$
• A Mathematical Rabbit out of an Algebraic Hat $\left[x^{2} + y^{2} + z^{2} - xy - yz - zx \ge 0\right]$
• An Inequality With an Infinite Series $\left[\displaystyle \sum_{n=1}^{1000}\frac{1}{n^{3}+3n^{2}+2n} \lt \frac{1}{4}\right]$
• An Inequality: $\left[1/2 \cdot 3/4 \cdot 5/6 \cdot \ldots \cdot 99/100 \lt 1/10\right]$
• $\left[\text{A Low Bound for }1/2 \cdot 3/4 \cdot 5/6 \cdot \ldots \cdot [2n-1]/2n\right]$
• An Inequality: Easier to prove a subtler inequality
• Inequality with Logarithms [logπ3 + log3π > 2]
• An inequality: 1 + 1/4 + 1/9 + ... less than 2
• Inequality with Harmonic Differences $\left[\displaystyle\frac{2n}{3n+1} \lt \frac{1}{n+1}+\frac{1}{n+2}+\ldots +\frac{1}{n+n}\lt \frac{25}{36}\right]$
• An Inequality by Uncommon Induction $\left[\displaystyle 1+\frac{1}{2^2}+\frac{1}{3^2}+\ldots +\frac{1}{n^2}\gt \frac{3n}{2n+1}\right]$
• Hlawka's Inequality $\left[|x+y|+|y+z|+|z+x|\le |x|+|y|+|z|+|x+y+z|\right]$
  • An Application of Hlawka's Inequality $\left[\sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2}+\sqrt{c^2-ca+a^2}\le a+b+c+\sqrt{a^2+b^2+c^2-ab-bc-ca}\right]$
  • An Application of Hlawka's Inequality II $\left[\displaystyle xPA+yPB+zPC+\frac{a\sqrt{3}}{2}PM\ge [y+z]PA'+[z+x]PB'+[x+y]PC'\right]$
  • Geometric Proof of Hlawka's Inequality $\left[|\vec{a}|+|\vec{b}|+|\vec{c}|+|\vec{d}|\ge |\vec{a}+\vec{d}|+|\vec{b}+\vec{d}|+|\vec{c}+\vec{d}|\right]$
  • Hlawka-like Dinca's Inequality $\left[[x+y]^{\alpha}+[y+z]^{\alpha}+[z+x]^{\alpha}\le x^{\alpha}+y^{\alpha}+z^{\alpha}+[x+y+z]^{\alpha}\right]$
  • Hlawka-like Inequality for Convex Functions $\left[g[x] + g[y] + g[z] + g[x + y + z] \ge g[x + y] + g[x + z] + g[y + z]\right]$
  • Hlawka in Convex Quadrilateral $\left[AB+BC+CD+DA\gt AC+BD+2EF\right]$
  • Quadrilateral Inequality
  • Problem 4165 from Crux Mathematicorum $\left[\displaystyle \left|\sum_{i=1}^4x_i\right|+2\sum_{i=1}^4|x_i|\ge 6\sqrt[6]{\prod_{1\le i\lt j\le 4}|x_i+x_j|}\right]$
• An Inequality in Determinants $\left[\left|\begin{array}{ccc} 1 & 1 & 1\\ a & b & c\\ a^4 & b^4 & c^4\\ \end{array}\right|\ge \left|\begin{array}{ccc} 1 & 1 & 1\\ a & b & c\\ a^3 & b^3 & c^3. \end{array}\right|\right]$
• Application of Cauchy-Schwarz Inequality $\left[\displaystyle\sqrt{a^2-1}+\sqrt{b^2-1}+\sqrt{c^2-1}\le\frac{ab+bc+ca}{2}\right]$
• An Inequality from Tibet $\left[[bcde+acde+abde+abce+abcd]^4\ge 125[a+b+c+d+e][abcde]^3\right]$
• An Inequality from Morocco $\left[\displaystyle 4|[a-b][b-c][c-d][d-a]|\le abcd\right]$
• An Inequality for Mixed Means $\small{\left[\displaystyle \sqrt[m]{\prod_{j=1}^{m}\left[a_j+\frac{1}{n}\sum_{i=1}^{n}x_i\right]}\ge\frac{1}{n}\sum_{i=1}^{n}\sqrt[m]{\prod_{j=1}^{m}[a_j+x_i]} \ge\prod_{i=1}^{n}\prod_{j=1}^{m}\sqrt[mn]{a_j+x_i}\right]}$
• An Inequality in Integers $\left[\displaystyle\frac{a}{b}+\frac{1}{ab}\lt\sqrt{7}\right]$
• An Inequality in Integers II $\left[\displaystyle\frac{a^3+b^3+c^3}{3}-abc\ge 3k\right]$
• An Inequality in Integers III $\left[\displaystyle\sum_{i,j=1}^n \frac{i^2j^2}{i+j-1}\gt n^3\right]$
• An Inequality with Exponents $\left[\displaystyle e^{\frac{4}{e}}\left[b\cdot a^{2\sqrt{a}}+c\cdot b^{2\sqrt{b}}+a\cdot c^{2\sqrt{c}}\right]\ge 3\sqrt[3]{abc}\right]$
• Exponential Inequalities for Means $\left[\displaystyle A_n^{G_nH_n}\le G_n^{A_nH_h}\le H_n^{A_nG_n}\right]$
• A Simple Inequality in Three Variables $\left[\displaystyle\sum_{cyc}ac\left[\frac{1}{2a+b}+\frac{1}{2c+b}\right]\le \sqrt{3[a^2+b^2+c^2]}\right]$
• An Asymmetric Inequality $\left[[a+b][a+c]\ge 2\sqrt{abc[a+b+c]}\right]$
• Linear Algebra Tools for Proving Inequalities $\left[\displaystyle\left[\frac{a}{b-c}\right]^2+\left[\frac{b}{c-a}\right]^2+\left[\frac{c}{a-b}\right]^2\ge 2.\right]$
• An Inequality with a Generic Proof $\left[2\sqrt{b}+\sqrt{aS+b}\le \sqrt{ax+b}+ \sqrt{ay+b}+ \sqrt{az+b}\le \sqrt{3aS+9b}\right]$
• A Generalization of an Inequality from a Romanian Olympiad $\left[\displaystyle \frac{2}{n-1}\sum_{1\le i\lt j\le n}a_{i}a_{j}\le\ln\left[\prod_{k=1}^{n}a_{k}^{a_{k}}\right]+\frac{1}{e}\sum_{k=1}^{n}e^{a_{k}}\right]$
• Area Inequality in Trapezoid $\left[[PQRS]\le\frac{1}{4}[ABCD]\right]$
• Improving an Inequality $\left[\displaystyle\sum_{k=1}^{n}\frac{1}{\sqrt{k}}\gt 2[\sqrt{n+1}-1]\right]$
• RomanoNorwegian Inequality $\left[\sqrt{\cot A} + \sqrt{\cot B} + \sqrt{\cot C} \ge 2\right]$
• Inequality with Nested Radicals II $\left[\displaystyle\sqrt{n+1}\le\sqrt{n+\sqrt{[n-1]+\sqrt{\ldots+\sqrt{2+\sqrt{1}}}}}\le\sqrt{2n-1}\right]$
• Inequality with Powers And Radicals $\left[\displaystyle\sum_{cycl}\sqrt[6]{ab^2c^3}\ge\sum_{cycl}\sqrt[30]{a^9b^{10}c^{11}}\right]$
• Inequality with Two Minima $\left[\displaystyle\sqrt{a-c}+\sqrt{b-c}\le\min \left\{\sqrt{\frac{ab}{c}},\sqrt{2[a+b-2c]} \right\}\right]$
• Simple Inequality with Many Faces And Variables $\left[\displaystyle \frac{a^2}{c-1}+\frac{b^2}{d-1}+\frac{c^2}{e-1}+\frac{d^2}{a-1}+\frac{e^2}{b-1}\ge 20\right]$
• An Inequality with Determinants $\left[\displaystyle\left|\begin{array}{cccc}\,a&-b&0&0\\0&b&-c&0\\0&0&c&-d\\1&1&1&1+d\end{array}\right|\ge 3\sqrt[3]{4}[abcd]^{5/6}\right]$
• An Inequality with Determinants II $\left[\displaystyle \Delta=\left|\begin{array}{cccc} \,1 & 1 & 1 & 1\\ a & b & c & d\\ a^2 & b^2 & c^2 & d^2\\ \frac{1}{a^2} & \frac{1}{b^2} & \frac{1}{c^2} & \frac{1}{d^2} \end{array}\right|\lt \frac{1}{abcd}\left[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right]\right]$
• An Inequality with Determinants III $\left[\displaystyle \left|\begin{array}{ccccc} \,0 & x^2 & y^2 & z^2 & 1\\ x^2 & 0 & x^2+y^2 & x^2+z^2 & 1\\ y^2 & x^2+y^2 & 0 & y^2+z^2 & 1\\ z^2 & x^2+z^2 & y^2+z^2 & 0 & 1\\ 1 & 1 & 1 & 1 & 0\end{array}\right|\le \left|\begin{array}{ccccc} \,0 & a^2 & b^2 & c^2 & 1\\ a^2 & 0 & a^2+b^2 & a^2+c^2 & 1\\ b^2 & a^2+b^2 & 0 & b^2+c^2 & 1\\ c^2 & a^2+c^2 & b^2+c^2 & 0 & 1\\ 1 & 1 & 1 & 1 & 0\end{array}\right|\right]$
• An Inequality with Determinants IV $\left[\displaystyle\left|\begin{array}{cccc}\,a&-b&0&0\\0&b&-c&0\\0&0&c&-d\\1&1&1&1+d\end{array}\right|\ge 3\sqrt[3]{4}[abcd]^{5/6}\right]$
• An Inequality with Determinants V $\left[\displaystyle \Delta =\left|\begin{array}{cccc} \,s & \frac{a^2b}{a^3+b} & \frac{b^2c}{b^3+c} & \frac{c^2a}{c^3+a}\\ \frac{a^2b}{a^3+b} & s & \frac{c^2a}{c^3+a} & \frac{b^2c}{b^3+c}\\ \frac{b^2c}{b^3+c} & \frac{c^2a}{c^3+a} & s & \frac{a^2b}{a^3+b}\\ \frac{c^2a}{c^3+a} & \frac{b^2c}{b^3+c} & \frac{a^2b}{a^3+b} & s\end{array}\right|\ge 0\right]$
• An Inequality with Determinants VI $\left[\displaystyle\frac{\Delta_1-\Delta_2}{[b-a][a-c][b-c]}\ge 12\sqrt[6]{[abc]^5}\right]$
• An Inequality with Determinants VII $\left[\displaystyle \left|\begin{array}{ccccc} \,0 & a^2 & b^2 & c^2 & 1\\ a^2 & 0 & a^2+b^2 & a^2+c^2 & 1\\ b^2 & a^2+b^2 & 0 & b^2+c^2 & 1\\ c^2 & a^2+c^2 & b^2+c^2 & 0 & 1\\ 1 & 1 & 1 & 1 & 0\end{array}\right| \le\frac{1}{8}\prod_{cycl}[a+b]^2\right]$
• An Inequality in Reciprocals $\left[\displaystyle\begin{align}\frac{4}{9}\left[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right]&+4\left[\frac{1}{a+2b}+\frac{1}{b+2c}+\frac{1}{c+2a}\right]\\ &\ge 5\left[\frac{1}{2a+b}+\frac{1}{2b+c}+\frac{1}{2c+a}\right]+\frac{1}{a+b+c}. \end{align}\right]$
• An Inequality in Reciprocals II $\left[\displaystyle \frac{1}{4a+1} + \frac{1}{4b+1} + \frac{6}{2a+2b+1} \ge \frac{4}{3a+b+1} + \frac{4}{3b+a+1}\right]$
• An Inequality in Reciprocals III $\left[\displaystyle \frac{1}{3}\sum_{cycl}\frac{1}{a+b} + \sum_{cycl}\frac{1}{4a+b+c}\ge\sum_{cycl}\left[\frac{1}{a+2b+3c}+\frac{1}{a+3b+2c}\right]\right]$
• A Problem from the Danubius Contest 2016 $\left[\displaystyle\{E[a,b]|\;a\gt 0, b\gt 0\}\subset \left[\frac{11}{10},\frac{10}{9}\right]\right]$
• A Problem from the Danubius-XI Contest $\left[\displaystyle\frac{a+b}{[1+c]^2}+\frac{b+c}{[1+a]^2}+\frac{c+a}{[1+b]^2}\ge\frac{9S}{[9+S]^2}\right]$
• An Inequality with Integrals and Rearrangement $\left[\displaystyle \sum_{cycl}b^2c^3\int_0^a[\cot x][\arctan x]dx\lt abc[a^3+b^3+c^3]\right]$
• An Inequality with Cot, Cos, and Sin $\left[\displaystyle\sum_{cycl}\cot^2A\,\cot^2B\ge\frac{\displaystyle\sum_{cycl}\cos^2A}{\displaystyle\sum_{cycl}\sin^2A}\right]$
• A Trigonometric Inequality from the RMM $\left[\displaystyle\sin [x+y]\le\sin x\left[\frac{\sin y}{y}\right]^3+\sin y\left[\frac{\sin x}{x}\right]^3\right]$
• An Inequality with Finite Sums $\left[\displaystyle\sum_{k=1}^n\frac{[2+x]^k}{2+x\cdot k}\ge 2^n-1\right]$
• Hung Viet's Inequality $\left[\displaystyle 1+\sum_{cycl}\frac{ab}{a^2-ab+b^2}\ge\frac{3}{2}\cdot\frac{[a+b][b+c][c+a]}{a^3+b^3+c^3}\right]$
• Hung Viet's Inequality II $\left[\displaystyle\min_{all}[a-b]^2\le\frac{a^2+b^2+c^2+d^2}{5}\right]$
• Hung Viet's Inequality III $\left[\displaystyle \left[\sum_{cycl}a^4\right]\left[\sum_{cycl}ab^3\right]\ge\left[\sum_{cycl}a^3b\right]\left[\sum_{cycl}a^2b^2\right]\right]$
• Inequality by Calculus $\left[\displaystyle\frac{b^{n+1}-a^{n+1}}{n+1}+\frac{ab[b^{n-1}-a^{n-1}]}{n-1}\leq [b-a]\sqrt{2[a^{2n}+b^{2n}]}\right]$
• Dorin Marghidanu's Calculus Lemma $\left[\displaystyle u^v\gt\frac{u}{u+v}\right]$
  • An Inequality with Exponents from a Calculus Lemma $\left[\displaystyle a^b\cdot [a+b]^c\cdot [a+b+c]^d\gt\frac{a}{a+b+c+d}\right]$
  • An Inequality with Exponents from a Calculus Lemma II $\left[2^a[b+c]^{1-a}+2^b[c+a]^{1-b}+2^c[a+b]^{1-c}\lt 4[a+b+c]\right]$
• An Area Inequality $\left[\displaystyle a^{2} + b^{2} + u^{2} + v^{2} + au + bv \ge \frac{3}{2}\right]$
• A 4-variable Inequality from the RMM $\left[\displaystyle |[a-b][b-c][c-a]|\le\sum_{cycl}|a-b||a+c+\eta||b+c+\eta|\right]$
• An Inequality from RMM with Powers of 2 $\left[2^x+2^y+2^z+2^{x+y+z}\gt\sqrt[x+y]{16^{xy}}+\sqrt[y+z]{16^{yz}}+\sqrt[z+x]{16^{zx}}+1\right]$
• A Cycling Inequality with Integrals $\bigg[2bc\Omega[a]+2ca\Omega[b]+2ab\Omega[c]\lt a^2+b^2+c^2$ where $\Omega[t]=\displaystyle\int_{0}^{1}\frac{1-x^2}{1+tx^2+x^4}dx\bigg]$
• A Cycling Inequality with Integrals II $\bigg[\Omega[a,b,c]+\Omega[b,c,a]+\Omega[c,a,b]\le 1,$ where $\Omega[p,q,r]=\displaystyle\int_{0}^{1}\frac{x^p}{1+x^p+x^{q+r}}dx\bigg]$
• An Inequality with Absolute Values $\left[\displaystyle\sum_{cycl}\frac{|a|+|b|}{1-c^2} \ge \sum_{cycl}\frac{2|a|}{1-bc}\right]$
• An Inequality from RMM with a Generic 5 $\left[\displaystyle\sqrt[3]{[2a+5][2b+5][2c+5]}\ge\frac{6abc}{ab+bc+ca}+5\right]$
• An Elementary Inequality by Non-elementary Means $\left[12[a\sin a + \cos a - 1]^2 \le 2a^4 + a^3\sin 2a\right]$
• Inequality in Quadrilateral $\left[\displaystyle\sum_{cycl}\sqrt{a^2+b^2+c^2}\gt 2\sqrt{3\cdot AC\cdot BD}\right]$
• Marian Dinca's Refinement of Nesbitt's Inequality $\left[\displaystyle\begin{align}\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}&\ge\frac{[a+b+c]^2}{2[ab+bc+ca]}\\&\ge\frac{3\sqrt{3[a^2+b^2+c^2]}}{2[a+b+c]}\\&\ge\frac{3}{2}\end{align}\right]$
• An Inequality in Cyclic Quadrilateral $\left[\displaystyle\frac{R}{8\sqrt{2}}\ge\frac{\sqrt[4]{[abcd]^3}}{[a+b+c+d]^2}\right]$
• An Inequality in Cyclic Quadrilateral II $\left[\displaystyle\sin A\sin B\le [\frac{s}{a}-1][\frac{s}{b}-1][\frac{s}{c}-1][\frac{s}{d}-1]\right]$
• An Inequality in Cyclic Quadrilateral III $\left[\displaystyle\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}\cos\frac{D}{2}\le\frac{S^2}{4abcd}\right]$
• An Inequality in Cyclic Quadrilateral IV $\left[a^2-b^2-c^2+d^2+4S\le 2\sqrt{2}[ad+bc]\right]$
• Inequality with Three Linear Constraints $\left[\displaystyle\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge 1\right]$
• Inequality with Three Numbers, Not All Zero $\left[\displaystyle\frac{a^2-ab+b^2}{b^2+bc+c^2}+\frac{b^2-bc+c^2}{c^2+ca+a^2}+\frac{c^2-ca+a^2}{a^2+ab+b^2}\ge 1\right]$
• An Easy Inequality with Three Integrals $\left[\displaystyle a^2\int_{0}^{b}\frac{\arctan x}{x}dx + b^2\int_{0}^{c}\frac{\arctan x}{x}dx + c^2\int_{0}^{a}\frac{\arctan x}{x}dx \lt a^3+b^3+c^3\right]$
• Divide And Conquer in Cyclic Sums $\left[\displaystyle \sum_{cycl}c\left[\frac{4a}{b^2}+\frac{3b}{a^2}\right]\ge 12+3\sum_{cycl}\frac{a}{b}\right]$
• Wu's Inequality $\left[[x^2+xy+y^2][y^2+yz+z^2][z^2+zx+x^2]\ge [xy+yz+zx]^3\right]$
• Dorin Marghidanu's Inequality in Complex Plane $\left[\displaystyle \sum_{cycl}[|[2-n]\cdot z_1+z_2+\ldots+z_n|\ge\sum_{k=1}^n|z_k|\right]$
• Dorin Marghidanu's Inequality in Integer Variables $\left[\displaystyle \frac{m}{\sqrt[m]{1+n}}+\frac{n}{\sqrt[n]{1+m}}\gt\frac{m+n}{2}\right]$
• Dorin Marghidanu's Inequality in Many Variables $\left[\displaystyle\prod_{k=1}^{n}\sqrt[n]{\frac{\displaystyle \sum_{i=1,i\ne k}^{k}a_i}{a_k]}} \ge n-1\right]$
• Dorin Marghidanu's Inequality in Many Variables Plus Two More $\left[\displaystyle G_n\left[p+\frac{r}{a_1},p+\frac{r}{a_2},\ldots,p+\frac{r}{a_n}\right]\ge p+\frac{r}{A_n[a_1,a_2,\ldots,a_n]}\right]$
• Dorin Marghidanu's Inequality with Radicals $\left[\displaystyle\sum_{k=1}^{n}\sqrt[i_k]{x_k} \gt \sqrt[\small{\displaystyle \sum_{k=1}^{n}i_k}]{\prod_{k=1}^{n}x_k}\right]$
• Dorin Marghidanu's Light Elegance in Four Variables $\left[\displaystyle \sum_{cycl}[-a+b+c+d]^2\ge 2[a+b+c+d]-1\right]$
• Dorin Marghidanu's Spanish Problem $\left[\displaystyle n^*\le [n_*]^2\right]$
• Two-Sided Inequality - One Provenance $\left[\displaystyle\sum_{k=1}^{2n[n+1]}\frac{1}{\sqrt{2k}+\sqrt{2k+1}}\lt n\lt\sum_{k=1}^{2n[n+1]}\frac{1}{\sqrt{2k-1}+\sqrt{2k}}\right]$
• An Inequality with Factorial $\left[a_1\cdot a_2\cdot\ldots\cdot a_n+[1-a_1]\cdot [2-a_2]\cdot\ldots\cdot [n-a_n]\le n!\right]$
• Wonderful Inequality on Unit Circle $\left[\displaystyle\left[\frac{a+b}{1+ab}\right]^2+\left[\frac{a-b}{1-ab}\right]^2\ge 1\right]$
• Quadratic Function for Solving Inequalities $\left[[a^2+3x^2][b^2+3y^2][c^2+3z^2]\ge 4[ayz+bzx+cxy+xyz]^2\right]$
• An Inequality Where One Term Is More Equal Than Others $\left[\displaystyle\left[\sum_{k=1}^na_k\right]\left[\sum_{k=1}^n\frac{1}{a_k}\right]\ge n^2+[n-2]^2\right]$
• Complicated Constraint - Simple Inequality $\left[3[a+b][b+c][c+a]\ge\frac{\displaystyle 8}{\displaystyle\sqrt[8]{a^3+b^3+c^3}}\right]$
• The power of substitution II: proving an inequality with three variables $\left[\displaystyle\frac{ab}{[a+b]^2}+\frac{bc}{[b+c]^2}+\frac{ca}{[c+a]^2}\le\frac{1}{4}+\frac{4abc}{[a+b][b+c][c+a]}\right]$
• Algebraic-Geometric Inequality $\left[\sqrt{x^2-\sqrt{3}xy+y^2} + \sqrt{y^2-\sqrt{2}yz+z^2} \ge \sqrt{z^2-zx+x^2}\right]$
• One Inequality - Two Domains $\left[\displaystyle 3\prod_{cycl}[a^2+ab+b^2]\ge\left[\sum_{cycl}a\right]^2\cdot\left[\sum_{cycl}ab\right]^2\right]$
• Radicals, Radicals, And More Radicals in an Inequality $\bigg[\displaystyle\gamma=\frac{\sqrt[4]{xz}}{\sqrt{x}+\sqrt{z}}.\,$ Prove that $\sqrt{x}+\sqrt{y}+\sqrt{z}\ge 2\gamma\sqrt{3[x+y+z]}\bigg]$
• An Inequality in Triangle and In General $\left[\displaystyle\sum_{cycl}\frac{\cot A\,\cot^3B}{\cot^2B+2\cot^2A}+2\sum_{cycl}\frac{\cot^2A\cot B}{\cot A+2\cot B}\ge 1\right]$
• Dan Sitaru's Cyclic Inequality In Many Variables $\left[\displaystyle a+b+c+d\le \frac{a^5+b^5+c^5+d^5}{abcd}\right]$
• An Inequality on Circumscribed Quadrilateral $\left[s\ge 4R\right]$
• An Inequality with Fractions $\left[\displaystyle m\le\frac{a_1+a_2+\ldots+a_n}{b_1+b_2+\ldots+b_n}\le M\right]$
• An Inequality with Complex Numbers of Unit Length $\left[|a-b|+|a-c|\ge |a+b|+|a+c|\right]$
• An Inequality with Complex Numbers of Unit Length II $\left[|a^2+bc|\ge |b+c|\right]$
• Le Khanh Sy's Problem $\left[xa^2+yb^2+zc^2\ge 2m\right]$
• An Inequality Not in Triangle $\left[\displaystyle\sqrt{a^2+b^2-ab\sqrt{2}}+\sqrt{b^2+c^2-bc\sqrt{3}}+\sqrt{c^2+d^2-\frac{cd[\sqrt{6}+\sqrt{2}]}{2}}\ge\sqrt{a^2+d^2}\right]$
• An Acyclic Inequality in Three Variables $\left[\displaystyle \frac{[a^2-bc]^2+[b^2-ca]^2+[c^2-ab]^2}{a^2+b^2+c^2+ab+bc+ca}\geq 3[a-b][b-c]\right]$
• An Inequality with Areas, Norms, and Complex Numbers $\left[\displaystyle \frac{[ad-bc][3[a^2+b^2][c^2+d^2]-4[ad-bc]^2]}{\left[[a^2+b^2][c^2+d^2]\right]^{\frac{3}{2}}}\le 1\right]$
• Darij Grinberg's Inequality In Three Variables $\left[a^2+b^2+c^2+2abc+1\ge 2[ab+bc+ca]\right]$
• Small Change Makes Big Difference $\left[\displaystyle\frac{1}{\displaystyle \sqrt{1+a^2-\frac{[a-b]^2}{2}}}+\frac{1}{\displaystyle \sqrt{1+b^2-\frac{[a-b]^2}{2}}}\ge\frac{2}{\sqrt{1+ab}}\right]$
• Inequality with Two Variables? Think Again $\left[\displaystyle\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}}\le\frac{2}{\sqrt{1+ab}}\right]$
• A Problem From a Mongolian Olympiad for Grade 11 $\left[\displaystyle \frac{a}{3a+2b^3}+ \frac{b}{3b+2c^3}+ \frac{c}{3c+2a^3}\le\frac{1}{5}\left[\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right]\right]$
• Sitaru--Schweitzer Inequality $\left[\displaystyle \left[\int_{a}^{b}f[x]dx\right]\left[\int_{a}^{b}\frac{1}{f[x]}dx\right]\le\frac{[m+M]^2}{4mM}[b-a]^2\right]$
  • Pólya-Szegö Inequality $\left[\displaystyle \frac{\displaystyle \left[\sum_{k=1}^{n}a^2_{k}\right]\left[\sum_{k=1}^{n}b^2_{k}\right]}{\displaystyle\left[\sum_{k=1}^{n}a_{k}b_{k}\right]^2}\le\left[\frac{\displaystyle \sqrt{\frac{M_1M_2}{m_1m_2}}+\sqrt{\frac{m_1m_2}{M_1M_2}}}{2}\right]^2\right]$
  • Kantorovich Inequality $\left[\displaystyle \left[\sum_{k=1}^{n}\gamma_ku_k^2\right]\left[\sum_{k=1}^{n}\frac{1}{\gamma_k}u_k^2\right]\le\frac{1}{4}\left[\sqrt{\frac{M}{m}}+\sqrt{\frac{m}{M}}\right]^2\left[\sum_{k=1}^{n}u^2_{k}\right]^2\right]$
  • Greub-Rheinboldt Inequality $\left[\displaystyle \left[\sum_{k=1}^{n}a_k^2u_k^2\right]\left[\sum_{k=1}^{n}b_k^2u_k^2\right]\le\frac{[M_1M_2+m_1m_2]^2}{4m_1m_2M_1M_2}\left[\sum_{k=1}^{n}a_kb_ku^2_{k}\right]^2\right]$
• An Inequality with Cyclic Sums And Products $\left[\small{\displaystyle \sum_{cycl}\frac{a^2}{[b+c+d+e][a-b][a-c][a-d][a-e]}\lt\frac{[a+b+c+d+e]^2}{1024abcde}}\right]$
• Problem 1 From the 2016 Pan-African Math Olympiad $\left[\displaystyle \sum_{cycl}\frac{1}{[x+1]^2+y^2+1}\le\frac{1}{2}\right]$
• An Inequality with Integrals and Radicals $\left[\displaystyle \Bigr[\int_0^1 \sqrt[3]{f[x]}dx\Bigr]\Bigr[\int_0^1 \sqrt[5]{f[x]}dx\Bigr]\Bigr[\int_0^1 \sqrt[7]{f[x]}dx\Bigr]\leq 1\right]$
• Twin Inequalities in Four Variables: Twin 1 $\left[\displaystyle [ac+bd]^2\le\left[b\sqrt[5]{ab^4}+d\sqrt[5]{cd^4}\right]\left[a\sqrt[5]{a^4b}+c\sqrt[5]{c^4d}\right]\right]$
• Twin Inequalities in Four Variables: Twin 2 $\left[\displaystyle [a\sqrt[3]{a^2b}+c\sqrt[3]{c^2d}][b\sqrt[3]{ab^2}+d\sqrt[3]{cd^2}]\le [a^2+c^2][b^2+d^2]\right]$
• Simple Inequality with a Variety of Solutions $\left[\displaystyle \sum_{cycl}\left[\frac{\ln x}{\ln y\ln z}+\frac{\ln y}{\ln z\ln x}\right]\ge\frac{18}{\ln [xyz]}\right]$
• A Partly Cyclic Inequality in Four Variables $\left[\displaystyle \sum_{cycl}xe^x\ge [x+y+2]e^{x+y+2}+[z+t-2]\sqrt[3]{e^{z+t-2}}\right]$
• Dan Sitaru's Inequality by Induction $\left[\displaystyle\begin{align}&\small{\frac{3}{a+1}+\frac{3}{b+1}+\frac{2}{c+1}+\frac{1}{d+1}}\\ &\small{\qquad\le 6+\frac{1}{a+b+1}+\frac{1}{a+b+c+1}+\frac{1}{a+b+c+d+1}}\end{align}\right]$
• An Inequality in Three [Or Is It Two] Variables $\left[\displaystyle \frac{[x+y]^2}{[x\sin^2z+y\cos^2z][x\cos^2z+y\sin^2z]}+\frac{x}{y}+\frac{y}{x}\geq 6\right]$
• An Inequality in Four Weighted Variables $\left[\displaystyle [a+c]^c[b+d]^d[c+d]^{c+d}\le c^cd^d[a+b+c+d]^{c+d}\right]$
• An Inequality in Fractions with Absolute Values $\left[\displaystyle \omega\lt\frac{1}{3}\left[\sum_{cycl}\frac{a|a|-b|b|}{a-b}\right]\lt 2\Omega\right]$
• Inequalities with Double And Triple Integrals $\left[\displaystyle \int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}}\cos\left[\frac{x+y}{2}\right]dxdy\ge\frac{\pi}{2}\right]$
• An Old Inequality $\left[\displaystyle \frac{4ab}{[a+b]^2}\ge\cos\left[\frac{\pi}{2}\cdot\frac{a-b}{a+b}\right]\right]$
• Dan Sitaru's Amazing, Never Ending Inequality $\left[\displaystyle \small{\sum_{cycl}\left[\frac{a}{b}\right]^2\cdot\sum_{cycl}\left[\frac{a}{b}\right]^4\cdot\sum_{cycl}\left[\frac{a}{b}\right]^8\ge\sum_{cycl}\left[\frac{a}{c}\right]\cdot\sum_{cycl}\left[\frac{b}{a}\right]\cdot\sum_{cycl}\left[\frac{b}{c}\right]}\right]$
• Leo Giugiuc's Exercise $\left[\displaystyle x\sin x+x^2\cos x\le 2\sin^2x\right]$
• Another Inequality with Logarithms, But Not Really $\left[\displaystyle \sum_{cycl}\frac{\log_y^3x+\log_z^3y}{\log_y^2x+\log_zx+\log_z^2y}\geq 2\right]$
• An Inequality Solved by Changing Appearances $\left[\displaystyle \sum_{cycl}a^2\cdot\sum_{cycl}\frac{1}{x^2}+\frac{\displaystyle 2\sum_{cycl}ab\cdot\sum_{cycl}x}{xyz}\ge 0\right]$
• Distances to Three Points on a Circle $\left[3\le |z-a|+|z-b|+|z-c|\le 4\right]$
• An Inequality with Powers And Logarithm $\left[\displaystyle \frac{a}{b}+\frac{a^2}{b^2}+\frac{a^3}{b^3}+12\ln b\ge\frac{b}{a}+\frac{b^2}{a^2}+\frac{b^3}{a^3}+12\ln a\right]$
• Four Integrals in One Inequality $\left[\displaystyle \small{\left[\int_a^bxf[x]dx\right]\left[\int_a^bf^2[x]dx\right]\left[\int_a^bx^3f[x]dx\right]\ge\frac{a^2b^2}{b-a}\left[\int_a^bf[x]dx\right]^4}\right]$
• Same Integral, Three Intervals $\left[\displaystyle\small{I[u,v]=\int_u^v\left[\arctan\left[\frac{u\sin x}{v+u\cos x}\right]+\arctan\left[\frac{v\sin x}{u+v\cos x}\right]\right]dx}\right]$
• Dorin Marghidanu's Inequality with Generalization $\left[\displaystyle [x+y]^2+[y+z]^2+[z+x]^2+12ab\le 4[a+b][x+y+z]\right]$
• Dan Sitaru's Inequality with Three Related Integrals and Derivatives $\left[\displaystyle\small{\left[\int_0^af[x]dx\right]^4\leq \frac{a^8}{60}\left[\int_0^a \left[f'[x]\right]^2 dx\right]\left[\int_0^a \left[f''[x]\right]^2dx\right]}\right]$
• An Inequality in Two Or More Variables $\left[\displaystyle \frac{a}{1+a}+\frac{b}{[1+a][1+b]}+\frac{c}{[1+a][1+b][1+c]}\geq \frac{7}{8}\right]$
• An Inequality in Two Or More Variables II $\left[\displaystyle [a+1]^{a+1}\cdot [b+1]^{b+1}\cdot [c+1]^{c+1}\le e^{a+b+c}\cdot\sqrt{e^{a^2+b^2+c^2}}\right]$
• A Not Quite Cyclic Inequality $\left[\displaystyle \frac{a^2+b^2+c^2}{a+b+c} \le \frac{ab+bc+ca}{a+b+c} + |a-b|+|b-c|\right]$
• Dan Sitaru's Inequality: From Three Variables to Many in Two Ways $\left[\displaystyle a+b+c\ge\frac{3}{2}\right]$
• An Inequality with Sines But Not in a Triangle $\left[\displaystyle \prod_{cycl}\Bigr[a^2\sin \frac{2\pi}{a}+[a+1]^2\sin \frac{2\pi}{a+1}\Bigr]\gt 2^{16}\right]$
• An Inequality with Angles and Integers $\left[\displaystyle k^2\tan \alpha +l^2\tan \beta \geq \frac{2kl}{\sin [\alpha+\beta]}-[k^2+l^2]\cot [\alpha+\beta]\right]$
• Sladjan Stankovik's Inequality In Four Variables $\left[\displaystyle 2\sum_{cycl}a^2-3\frac{\displaystyle \sum_{cycl}a^3}{\displaystyle \sum_{cycl}a}\le\sum_{all}ab\right]$
• An Inequality with Two Pairs of Triplets $\left[\displaystyle [a^2+b^2+c^2]\left[\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right]+\frac{2[ab+bc+ca][x+y+z]}{xyz}\ge 0\right]$
• A Refinement of Turkevich's Inequality $\left[\displaystyle a^2+b^2+c^2+d^2+\frac{32abcd}{[a+b+c+d]^2}\ge \sum_{sym}ab\right]$
• Dan Sitaru's Exercise with Pi and Ln $\left[\displaystyle \small{\arctan[z-x]+\arctan[z-y]+\arctan[y-x] \lt \frac{\pi}{2} - \ln[2]}\right]$
• Leo Giugiuc's Cyclic Quickie in Four Variables $\left[3[xyz+yzt+ztx+txy]^2\ge 8[xy+xz+xt+yz+yz+zt]xyzt\right]$
• Dan Sitaru's Cyclic Inequality in Four Variables $\left[\displaystyle \sum_{cycl}\frac{a^7}{bcd+a^3}\ge 2abcd\right]$
• A Not Quite Cyclic Inequality from Tibet $\left[[x+y]^2[z^2+xz+x^2+xy+y^2+yz]^2\ge 8[xy+yz+zx]^2[x^2+y^2]\right]$
• An inequality in 2+2 variables from SSMA magazine $\left[\displaystyle k^2\tan\alpha+l^2\tan\beta\ge\frac{2kl}{\sin [\alpha+\beta]}-[k^2+l^2]\cot[\alpha+\beta]\right]$
• Kunihiko Chikaya's Inequality with Parameter $\bigg[p \ge 2.$ Prove $\displaystyle \sum_{cycl}\frac{a}{\sqrt{ap+b}} \le \sqrt{\frac{3[a+b+c]}{p+1}}\bigg]$
• Dorin Marghidanu's Permuted Inequality $\left[\displaystyle \sum_{k=1}^n\left[a_k+\frac{1}{a_{\sigma[k]}}\right]^p\ge \frac{[s^2+n^2]^p}{n^{p-1}s^p}\right]$
• An Inequality Involving Arithmetic And Geometric Means $\left[\displaystyle\sum_{cycl}\frac{1}{a^4+b^4+c^4+abcd}\le \frac{1}{abcd}\right]$
• Dorin Marghidanu's Sums and Products $\left[\displaystyle \sum_{k=1}^n\frac{a_k}{P_kS_k}\ge\frac{n^n}{\displaystyle [n-1]S^{n-1}}\right]$
• Simple Nameless Inequality $\left[\displaystyle \sum_{k=1}^n\frac{S}{S_k}\ge\frac{n^2}{n-1}\right]$
• Volume Inequality in Tetrahedron $\left[OA\cdot OB\cdot OC\ge 27xyz\right]$
• Inequality in Convex Quadrilateral $\left[\displaystyle\frac{\displaystyle \sum_{cycl}\sqrt{b+c+d-a}}{a+b+c+d}\ge\sqrt{\frac{2[a+b+c+d]}{a^2+b^2+c^2+d^2}}\right]$
• Dan Sitaru's Inequality with a Double Integral $\left[\displaystyle\begin{align}&\small{\int_0^1\int_0^1\sqrt{\left[m^2\sqrt{mnf[x]f[y]}+f^2[x]\right]\left[n^2\sqrt{mnf[x]f[y]}+f^2[y]\right]}dxdy}\\ &\qquad\qquad\qquad\qquad\small{\ge [m+n]\int_0^1f[x]dx}.\end{align}\right]$
• Cute Exercise by Dorin Marghidanu $\left[\displaystyle \sum_{k=1}^n\frac{2k-1}{\sqrt[2k-1]{a_k}}\ge\frac{n^2}{\sqrt[n^2]{a_1a_2\ldots a_n}}\right]$
• A Little of Algebra for an Inequality, A Little of Calculus for a Generalization $\left[\displaystyle \frac{a^{n+1}-b^{n+1}}{a-b}\cdot\frac{b^{n+1}-c^{n+1}}{b-c}\cdot\frac{c^{n+1}-a^{n+1}}{c-a}\gt [n+1]^3[abc]^n\right]$
• An Inequality with Central Binomials $\left[\displaystyle \sqrt{2}\le\sqrt[n[n+1]]{{2\choose 1}{4\choose 2}\cdots{2k\choose k}\cdots{2n\choose n}}\lt 2\right]$
• A Simple Inequality with Many Variables $\left[\displaystyle \sum_{k=1}^n\sqrt{\frac{a_k+a_{k+1}}{a_{k+2}}}\ge n\sqrt{2}\right]$
• Cyclic Inequality in Four Variables $\left[\displaystyle \frac{3}{4}\sum_{cycl}\frac{a^3}{bcd}\ge 1+\frac{\displaystyle 3\sum_{cycl}a^2}{\displaystyle \sum_{all}ab}\right]$
• Cyclic Inequality in Four Variables By D. Sitaru $\left[\displaystyle \sum_{cycl}\frac{a^7}{a^3+bcd}\ge 2abcd\right]$
• Lorian Saceanu's Inequality with Many Variables $\left[\displaystyle \frac{1}{2}\left[\sqrt{ab}+\frac{1}{\sqrt{ab}}\right]\left[\sum_{i=1}^na_kb_k\right]\ge\sqrt{\left[\sum_{i=1}^na_k^2\right]\left[\sum_{i=1}^nb_k^2\right]}\right]$
• A True Algebraic-Geometric Inequality $\left[\displaystyle \small{\sqrt{\sum_{k=1}^n[2a_k-b_k]^2}+\sqrt{\sum_{k=1}^n[2b_k-a_k]^2}\ge\sqrt{\sum_{k=1}^na_k^2}+\sqrt{\sum_{k=1}^nb_k^2}}\right].$
• Leo Giugiuc's Cyclic Inequality in Square Roots $\left[\displaystyle \sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x}\ge 2\sqrt{\frac{[x+y][y+z][z+x]}{xy+yz+zx}}\right]$

|Contact| |Front page| |Contents| |Up|

Copyright © 1996-2018 Alexander Bogomolny