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# Inequality 27(Vo Quoc Ba Can)

Problem:

Let $\displaystyle a,b,c$ be positive real numbers. Prove that

$\displaystyle \frac{a^{b+c}}{(b+c)^2}+\frac{b^{c+a}}{(c+a)^2}+\frac{c^{a+b}}{(a+b)^2}\geq \frac{3}{4}$

Solution:

It holds for every positive number $\displaystyle x$ that

$\displaystyle \frac{1}{x^2}\geq \frac{1}{x^{2x}}$.

Back to our inequality now, rewrite it as follows:

$\displaystyle \frac{a^{b+c}}{(b+c)^2}+\frac{b^{c+a}}{(c+a)^2}+\frac{c^{a+b}}{(a+b)^2}\geq \frac{3}{4}\Longrightarrow \sum_{cyc}\frac{4a^{b+c}}{(b+c)^2}\geq 3\Longrightarrow \sum_{cyc}\frac{a^{b+c}}{\left(\frac{b+c}{2}\right)^{2}}\geq 3$.

Using the above result we have that

$\displaystyle \sum_{cyc}\frac{a^{b+c}}{\left(\frac{b+c}{2}\right)^{2}}\geq \sum_{cyc}\left(\frac{a}{\frac{b+c}{2}}\right)^{b+c}=\sum_{cyc}\left(\frac{2a}{b+c}\right)^{b+c}$.

Let us now make the final step for the proof of our inequality. From Bernulli’s inequality we get that

$\displaystyle \sum_{cyc}\left(\frac{2a}{b+c}\right)^{b+c}=\sum_{cyc}\left[\left(\frac{2a}{b+c}-1\right)+1\right]^{b+c}\geq \sum_{cyc}(1+2a-b-c)=3$, Q.E.D.

## 2 thoughts on “Inequality 27(Vo Quoc Ba Can)”

1. Paolo Perfetti says:

Dear George,

in problem 32, are you sure that
$\displaystyle \frac{1}{1+a}+\frac{1}{1+b}- \frac{2}{1+\sqrt{ab}}\geq 0.$ Isn’t the quantity negative by taking, for instance, $\displaystyle a=1, b=\frac{1}{4}$?

• Well, you are right. I just forgot to mention that $\displaystyle abcd=1$ and $\displaystyle a\geq b\geq c\geq d>0$ imply that $\displaystyle \sqrt{ab}\geq 1$