# An Inequality via Young's Theorem

Dan Sitaru posted an elegant inequality and its proof at the CutTheKnotMath facebook page.

For $a,b,c\gt 0$ the following inequality holds

$\displaystyle ab+bc+ca\le\ln\left(a^a\cdot b^b\cdot c^c\right)+\frac{e^a+e^b+e^c}{e}.$

The equality is reached for $a=b=c=1.$

The above inequality follows from Young's theorem:

Let $f$ be a continuous strictly increasing function on an interval $[0,c]$ and $f^{-1}$ it inverse. For $a\in [a,c],$ and $b\in [0,f(c)],$

$\displaystyle ab\le \int_{0}^{a}f(x)dx+\int_{0}^{b}f^{-1}(x)dx,$

with the equality is only achieved if $b=f(a).$

### Proof of Sitaru's inequality

Choose $f(x)=\ln (x+1)$ so that $f^{-1}(x)=e^{x}-1.$ Obviously such $f$ satisfies the conditions of Young's inequality for any $c\gt 0.$

Now by elementary calculus

$\displaystyle \int_{0}^{a}f(x)dx=\int_{0}^{a}\ln (x+1)dx=(a+1)\ln (a+1)-a,$

and also

$\displaystyle \int_{0}^{b}f^{-1}(x)=\int_{0}^{b}(e^x-1)dx=e^b-1-b.$

By Young's theorem it follows that

$ab\le (a+1)\ln (a+1)-a + e^b-1-b.$

In other words,

$(a+1)(b+1)\le (a+1)\ln (a+1)+e^b.$

By replacing $a+1$ with $a$ and $b+1$ with $b$ we obtain

$\displaystyle ab\le a\ln a+e^{b-1}=\ln a^a+\frac{e^b}{e}.$

Similarly,

$\displaystyle bc\le \ln b^b+\frac{e^c}{e}$ and $\displaystyle ca\le \ln c^c+\frac{e^a}{e}.$

Adding the three proves the desired inequality.

Clearly, the statement and the proof extend to any number of parameters:

For positive $x_{1},x_{2},\ldots,x_{n},$ $n\ge 3,$ $x_{n+1}=x_{1},$

$\displaystyle \sum_{k=1}^{n}x_{k}x_{k+1}\le \ln\prod_{k=1}^{n} x_{k}^{x_{k}}+\frac{\displaystyle \sum_{k=1}^{n}e^{x_{k}}}{e}.$

Copyright © 1996-2017 Alexander Bogomolny

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