An Inequality in Triangle, with Sides and Sums

Problem

An Inequality in Triangle, with  Sides and Sums

Solution 1

Rewrite the inequality as

$\displaystyle \sum_{cycl}\frac{a(b+c)}{2(b+c-a)}\ge\sum_{cycl}a.$

Using Bergstrom's inequality,

$\displaystyle\begin{align}\sum_{cycl}\frac{a(b+c)}{b+c-a}&=\sum_{cycl}\left[\frac{a^2}{b+c-a}+a\right]\\ &\ge\frac{\displaystyle \left(\sum_{cycl}a\right)^2}{\displaystyle \sum_{cycl}(b+c-a)}+\sum_{cycl}a\\ &=\frac{\displaystyle \left(\sum_{cycl}a\right)^2}{\displaystyle \sum_{cycl}a}+\sum_{cycl}a\\ &= 2\sum_{cycl}a. \end{align}$

This is the required inequality.

Solution 2

It's easy to verify that for $\displaystyle f(x)=\frac{x(2s-x)}{4(s-x)},\,$ $f''(x)\gt 0\,$ so that the function is convex and Jensen's inequality applies: $\displaystyle f\left(\frac{a+b+c}{3}\right)\le\frac{1}{3}(f(a)+f(b)+f(c)).\,$ Now

$\displaystyle \begin{align} f\left(\frac{a+b+c}{3}\right) &= \frac{a+b+c}{3\cdot 4}\left(\frac{\displaystyle (a+b+c)-\frac{a+b+c}{3}}{\displaystyle \frac{a+b+c}{2}-\frac{a+b+c}{3}}\right)\\ &=\frac{a+b+c}{12}\cdot\frac{\displaystyle \frac{2(a+b+c)}{3}}{\displaystyle \frac{a+b+c}{6}}=\frac{a+b+c}{3}\\ &\le\frac{1}{3}(f(a)+f(b)+f(c)), \end{align}$

giving the required inequality.

Acknowledgment

The problem from the Romanian Mathematical Magazine has been kindly posted by Dan Sitaru at the CutTheKnotMath facebook page. Solution 2 is by Mehmet Sahin.

 

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