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# Inequality 56 (George Basdekis)

Problem:

Let $a,b,c$ be positive real numbers. Prove the Inequality

$\displaystyle \frac{a\sqrt{a}}{2a+b}+\frac{b\sqrt{b}}{2b+c}+\frac{c\sqrt{c}}{2c+a}\leq\frac{(a+b+c)\sqrt{a+b+c}}{3\sqrt{3}\sqrt[3]{abc}}.$

Solution:

First, we observe that from the AM-GM Inequality we have that

$\displaystyle 3a(a+2b)\leq\left[\frac{4a+2b}{2}\right]^2=(2a+b)^2$

or equivalently

$\displaystyle \frac{3a}{(2a+b)^2}\leq\frac{1}{a+2b}\Leftrightarrow \frac{3a^3}{(2a+b)^2}\leq\frac{a^2}{a+2b}$

and now taking square roots for both sides, we see that

$\displaystyle \sqrt{3}\cdot\frac{a\sqrt{a}}{2a+b}\leq\frac{a}{\sqrt{a+2b}}.$

Therefore, we acquire

$\displaystyle \sqrt{3}\sum_{cyc}\frac{a\sqrt{a}}{2a+b}\leq\sum_{cyc}\frac{a}{\sqrt{a+2b}}.$

Now, from the Cauchy-Schwarz Inequality we have for the right hand side

\displaystyle \begin{aligned}\left(\sum_{cyc}\frac{a}{\sqrt{a+2b}}\right)^2&\leq\sum_{cyc}[a(a+2c)]\sum_{cyc}\frac{a}{(a+2b)(a+2c)}\\&=(a+b+c)^2\sum_{cyc}\frac{a}{(a+2b)(a+2c)}.\end{aligned}

Moreover, once again, from the AM-GM Inequality we have for the denominators of the above sum that

$\displaystyle (a+2b)(a+2c)\geq 9\sqrt[3]{a^2b^2c^2}$

and thus, it will hold that

\displaystyle \begin{aligned}(a+b+c)^2\sum_{cyc}\frac{a}{(a+2b)(a+2c)}&\leq (a+b+c)^2\sum_{cyc}\frac{a}{9\sqrt[3]{a^2b^2c^2}}\\&=\frac{(a+b+c)^3}{9\sqrt[3]{a^2b^2c^2}}.\end{aligned}

Finally, we have proved that

$\displaystyle \left(\sum_{cyc}\frac{a}{\sqrt{a+2b}}\right)^2\leq\frac{(a+b+c)^3}{9\sqrt[3]{a^2b^2c^2}}\Leftrightarrow \sum_{cyc}\frac{a}{\sqrt{a+2b}}\leq\frac{(a+b+c)\sqrt{a+b+c}}{3\sqrt[3]{abc}}.$

The Inequality we have proved is

$\displaystyle \sqrt{3}\cdot\sum_{cyc}\frac{a\sqrt{a}}{2a+b}\leq\frac{(a+b+c)\sqrt{a+b+c}}{3\sqrt[3]{abc}}$

which also is the Inequality we are given to prove. The proof is completed Q.E.D.