A Generalization of an Inequality from a Romanian Olympiad

On an earlier page we discussed a problem from a Romanian olympiad and suggested its obvious generalization. Leo Giugiuc and Dan Sitaru proved a more involved generalization:

Proof

By MacLaurin's inequality,

$\displaystyle\frac{2}{n(n-1)}\sum_{1\le i\lt j\le n}a_{i}a_{j}\le\left(\frac{1}{n}\sum_{k=1}^{n}a^k\right)^2.$

On the other hand, by Jensen's inequality,

$\displaystyle\left(\frac{1}{n}\sum_{k=1}^{n}a_k\right)^2\le\frac{1}{n}\sum_{k=1}^{n}a_{k}^2.$

It follows that $\displaystyle\frac{2}{n-1}\sum_{1\le i\lt j\le n}a_{i}a_{j}\le\sum_{k=1}^{n}a_k^2.\;$ Thus, suffice it to prove that

$\displaystyle\sum_{k=1}^{n}a_{k}^2\le\ln\left(\prod_{k=1}^{n}a_{k}^{a_{k}}\right)+\frac{1}{e}\sum_{k=1}^{n}e^{a_{k}}.$

We'll use the following

Lemma

Proof of Lemma

$f'(x)=e^{x-1}+\ln x+1-2x\;$ and $\displaystyle f''(x)=e^{x-1}+\frac{1}{x}-2.\;$ By the AM-GM inequality, $\displaystyle e^{x-1}+\frac{1}{x}\ge 2\sqrt{\frac{e^{x-1}}{x}},\;$ which, combined with the well known fact that $e^{x-1}\ge (x-1)+1=x,\;$ for $x\gt 0,\;$ gives $\displaystyle e^{x-1}+\frac{1}{x}\ge 2,\;$ and implying that, for $x\gt 0,\;$ $f''(x)\ge 0,\;$ making $f'(x)\;$ increasing.

Since $f'(x)=0,\;$ we deduce that $f(x)\;$ is decreasing on $(0,1]\;$ and increasing on $[1,\infty )$ such that it achieves its minimum at $x=1.\;$ Therefore, $f(x)\ge f(1)=0.$

 

Back to the main problem, we apply the lemma for all $a_k\;$ and, adding up the $n\;$ inequalities up, obtain

$\displaystyle\ln\left(\prod_{k=1}^{n}a_{k}^{a_k}\right)+\frac{1}{e}\sum_{k=1}^{n}e^{a_k}-\sum_{k=1}^{n}a_{k}^2\ge 0,\;$ from which $\displaystyle\ln\left(\prod_{k=1}^{n}a_{k}^{a_k}\right)+\frac{1}{e}\sum_{k=1}^{n}e^{a_k}\ge\frac{2}{n-1}\sum_{1\le i\lt j\le n}a_{i}a_{j}.$

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• 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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