Inequality

An inequality proposed by one of our readers.

Let $a, b, c >0$. Prove that:

$$ \frac{ab}{a+b+2c}+ \frac{bc}{b+c+2a}+ \frac{ca}{c+a+2b}\leq \frac{a+b+c}{4}$$

Solution



Let $x, y>0$. From the AM-HM inequality we have that:
$$\frac{2}{\frac{1}{x}+ \frac{1}{y}}\leq \frac{x+y}{2} \tag{1}$$

Hence letting $x=\frac{1}{a+c}$ and $y=\frac{1}{b+c}$  we note that $(1)$ takes its equivalent form:

$$\begin{aligned} \frac{2}{(a+c)+(b+c)}\leq \frac{1}{2}\left ( \frac{1}{a+c}+ \frac{1}{b+c} \right )&\Rightarrow \frac{1}{a+b+2c}\leq \frac{1}{4}\left ( \frac{1}{a+c}+ \frac{1}{b+c}
 \right )\\
&\Rightarrow \frac{ab}{a+b+2c}\leq \frac{ab}{4}\left ( \frac{1}{a+c}+ \frac{1}{b+c} \right ) \end{aligned}$$

Doing the same procedure we get that:

$$\frac{bc}{b+c+2a}\leq \frac{bc}{4}\left ( \frac{1}{b+a}+ \frac{1}{c+a} \right )  \;\; {\rm and} \;\; \frac{ca}{c+a+2b}\leq \frac{ca}{4}\left ( \frac{1}{c+b}+ \frac{1}{a+b} \right )$$

Adding the three equations together we have that:

$$\begin{aligned}
\sum_{cyc}\frac{ab}{a+b+2c} &\leq \frac{1}{4}\left [ \frac{ab}{a+c}+ \frac{ab}{b+c}+ \frac{bc}{b+a}+ \frac{bc}{c+a} + \frac{ca}{c+b}+ \frac{ca}{a+b} \right ] \\
 &= \frac{1}{4}\left [ \left ( \frac{ab}{a+c}+ \frac{bc}{a+c} \right )+ \left ( \frac{ab}{b+c}+ \frac{ca}{b+c} \right )+ \left ( \frac{bc}{b+a} + \frac{ca}{a+b} \right ) \right ]\\
 &= \frac{1}{4}\left ( a+b+c \right )
\end{aligned}$$

which is the wanted result.