Simple Inequality with Many Faces And Variables

Inequality from India Regional MO 2013

Proof 1

By the Cauchy-Schwarz inequality,

$\displaystyle\begin{align} (a+b+&c+d+e)^2\\ &\le\left(\frac{a^2}{c-1}+\frac{b^2}{d-1}+\frac{c^2}{e-1}+\frac{d^2}{a-1}+\frac{e^2}{b-1}\right)(c+d+e+a+b-5). \end{align}$

In other words,

$\displaystyle \frac{a^2}{c-1}+\frac{b^2}{d-1}+\frac{c^2}{e-1}+\frac{d^2}{a-1}+\frac{e^2}{b-1}\ge\frac{(a+b+c+d+e)^2}{a+b+c+d+e-5}=\mathbf{T}.$

Suffice it to prove that $\mathbf{T}\ge 20,\;$ but this is equivalent to $(a+b+c+d+e-10)^2\ge 0\;$ which is obvious.

Proof 2

Note that $\displaystyle\frac{x}{2}=\frac{x-1+1}{2}\ge\sqrt{x-1}\;$ so that

$\displaystyle\begin{align} \frac{a^2}{c-1}+\frac{b^2}{d-1}&+\frac{c^2}{e-1}+\frac{d^2}{a-1}+\frac{e^2}{b-1}\\ &\ge\frac{a^2}{\displaystyle\left(\frac{c}{2}\right)^2}+\frac{b^2}{\displaystyle\left(\frac{d}{2}\right)^2}+\frac{c^2}{\displaystyle\left(\frac{e}{2}\right)^2}+\frac{d^2}{\displaystyle\left(\frac{a}{2}\right)^2}+\frac{e^2}{\displaystyle\left(\frac{b}{2}\right)^2}\\ &=4\sum_{cycl}\frac{a^2}{c^2}\\ &\ge4\cdot 5\sqrt[5]{\prod_{cycl}\frac{a^2}{c^2}}\\ &=20. \end{align}$

Equality is achieved for $a=b=c=d=e=2.$

Proof 3

Introduce $x=a-1,\;$ $y=b-1,\;$ $z=c-1,\;$ $u=d-1,\;$ $v=e-1.\;$ The required inequality becomes

$\displaystyle\frac{(x+1)^2}{z}+\frac{(y+1)^2}{u}+\frac{(z+1)^2}{v}+\frac{(u+1)^2}{x}+\frac{(v+1)^2}{y}\ge 20.$

The left-hand side opens to

$\displaystyle\left(\frac{x^2}{z}+\frac{x}{z}+\frac{x}{z}+\frac{1}{z}\right)+\left(\frac{y^2}{u}+\frac{y}{u}+\frac{y}{u}+\frac{1}{u}\right)+\ldots = \mathbf{T}.$

By the AM-GM inequality, $\displaystyle T\ge 20\sqrt[20]{\prod_{cycl}\frac{x^4}{z^4}}=20.$

Proof 4

By C-B-S, $\displaystyle\mbox{l.h.s.}\ge\frac{x^2}{x-5}\ge 20\;$ which is equivalent to $(x-10)^2\ge 0\;$ where $x=a+b+c+d+e.$

Proof 5

Let $n\ge 1,\;$ an integer, and $\tau\;$ a permutation of the set $\{1,2,\ldots,n\}.\;$ Then, for any $x_k\gt 1,\;$ $k=1,2,\ldots,n,\;$ we have $\displaystyle\sum_{k=1}^{n}\frac{x_k^2}{x_{\tau(k)}-1}\ge 4n.$

Indeed, assuming $\{x_k\}\;$ is monotone, $\displaystyle\frac{1}{x_k-1}\;$ is inversely monotone so that, by the Rearrangement inequality,

$\displaystyle\begin{align} \sum_{k=1}^{n}\frac{x_k^2}{x_{\tau(k)}-1} &\ge \sum_{k=1}^{n}\frac{x_k^2}{x_{k}-1}\\ &=\sum_{k=1}^{n}\frac{[(x_k-1)+1]^2}{x_{k}-1}\\ &=\sum_{k=1}^{n}\left((x_k-1)+2+\frac{1}{x_{k}-1}\right)\\ &\ge\sum_{k=1}^{n}(2+2)\\ &=4n, \end{align}$

by the AM-GM inequality. Equality is achieved when all $x_k-1\;$ are equal $1.$

Proof 6

Observe that $\displaystyle\frac{a^2}{c-1}+4(c-1)\ge 4a.\;$ Summing up gives the required inequality.

Acknowledgment

The problem has been posted by Diego Alvariz at Imad Zak Math facebook group and reproduced here with Imad Zak's kind permission. Proof 1 is by Kunihiko Chikaya; Proof 2 by Diego Alvariz; Proof 3 by Soumava Chakraborty; Proof 4 is Imad Zak's elucidation on Proof 1; Proof 5 is by Marian Dinca; Proof 6 is by .

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