# A Simple Inequality in Three Variables

### Problem

Dan Sitaru has kindly posted a problem at the CutTheKnotMath facebook page, followed by a solution (Solution 1) of his. Solution 4 is by Ghimisi Dumitrel.

If $a,b,c\gt 0,\;$ prove

$\displaystyle\sum_{cyc}ac\left(\frac{1}{2a+b}+\frac{1}{2c+b}\right)\le \sqrt{3(a^2+b^2+c^2)}.$

### Solution 1

First, let's check that, for $a,b\gt 0,\;$

$\displaystyle\frac{a}{2a+b}+\frac{b}{2b+a}\lt 1.$

Indeed, this is equivalent to $a(2b+a)+b(2a+b)\lt (2b+a)(2a+b),\;$ which simplifies to $0\lt a^2+b^2+ab\;$ and is, therefore, true.

Similarly, $\displaystyle\frac{b}{2b+c}+\frac{c}{2c+b}\lt 1\;$ and $\displaystyle\frac{c}{2c+a}+\frac{a}{2a+c}\lt 1.\;$ Multiplying these by $c,\;$ $a,\;$ $b,\;$ respectively,

$\displaystyle\frac{ac}{2a+b}+\frac{bc}{2b+a}\lt c,\\ \displaystyle\frac{ab}{2b+c}+\frac{ac}{2c+b}\lt a,\\ \displaystyle\frac{bc}{2c+a}+\frac{ba}{2a+c}\lt b.$

$\displaystyle\sum_{cyc}ac\left(\frac{1}{2a+b}+\frac{1}{2c+b}\right)\lt a+b+c \le\sqrt{3(a^2+b^2+c^2)}.$

### Solution 2

This solution is just a shortcut to Solution 1:

\displaystyle\begin{align} \sum_{cyc}ac\left(\frac{1}{2a+b}+\frac{1}{2c+b}\right)&=\sum_{cyc}\left(c\frac{a}{2a+b}+a\frac{c}{2c+b}\right)\\ &\lt\sum_{cyc}\left(c\frac{1}{2}+a\frac{1}{2}\right)\\ &=\frac{1}{2}\sum_{cyc}(a+c)\\ &=a+b+c\\ &\le\sqrt{3(a^2+b^2+c^2)}, \end{align}

because

$\displaystyle\frac{a+b+c}{3}\le\sqrt{\frac{a^2+b^2+c^2}{3}}.$

### Solution 3

Here we shall prove a strengthened inequality

If $a,b,c\gt 0,\;$ prove

Define function $f(x)=\displaystyle\frac{1}{2+x}+\frac{x}{2x+1}.\;$ $f'(x)=\displaystyle\frac{1-x^2}{(2+x)^2(2x+1)^2},\;$ implying that

$\displaystyle\begin{cases} f'(x)\gt 0,& 0\lt x\lt 1\\ f'(x)= 0,& x=1\\ f'(x)\lt 0,& x \gt 1. \end{cases}$

Thus $f(x)\;$ has a maximum at $x=1:\;$ $f(1)=\displaystyle\frac{2}{3}.\;$ It follows that

$\displaystyle\frac{ac}{2a+b}+\frac{bc}{2b+a}\lt \frac{2}{3}c,\\ \displaystyle\frac{ab}{2b+c}+\frac{ac}{2c+b}\lt \frac{2}{3}a,\\ \displaystyle\frac{bc}{2c+a}+\frac{ba}{2a+c}\lt \frac{2}{3}b,$

or,

$\displaystyle\frac{3}{2}\sum_{cyc}ac\left(\frac{1}{2a+b}+\frac{1}{2c+b}\right)\lt a+b+c \le\sqrt{3(a^2+b^2+c^2)}.$

### Solution 4

We'll prove first that, for $u,v\gt 0,$

$\displaystyle\frac{u}{2u+v}+\frac{v}{2v+u}\le\frac{2}{3}.$

Indeed, stating with $u^2-2uv+v^2\ge 0,$ we get

\begin{align} 3(u(2v+u)+v(2u+v)&=3(2uv+u^2+2uv+v^2)\\ &\le 2(4uv+2v^2+2u^2+uv)\\& =2(2u+v)(2v+u). \end{align}

Given that,

\displaystyle\begin{align} \frac{3}{2}\sum_{cyc}ac\left(\frac{1}{2a+b}+\frac{1}{2c+b}\right)&=\frac{3}{2}\sum_{cyc}a\left(\frac{b}{2b+c}+\frac{c}{2c+b}\right)\\ &\le\frac{3}{2}\sum_{cyc}a\frac{2}{3}\\ &=a+b+c\\ &\le\sqrt{3(a^2+b^2+c^2)}. \end{align}