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

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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).

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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)

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• 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 $\left(\displaystyle \sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x}\ge 2\sqrt{\frac{(x+y)(y+z)(z+x)}{xy+yz+zx}}\right)$

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