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Inequality 27(Vo Quoc Ba Can)


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}


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.


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