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# Inequality 42(George Basdekis)

Problem:

If $\displaystyle a,b,c$ are positive real numbers such that $\displaystyle abc=1$, then prove that

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

Solution:

From the Cauchy-Schwarz inequality we deduce that

$\displaystyle (a^3+b^3)(a+b)\geq (a^2+b^2)^2$.

Removing the square we get that

$\displaystyle \sqrt{a^3+b^3}\cdot\sqrt{a+b}\geq (a^2+b^2)$. Let us now divide by $\displaystyle a^2+b^2$.

Then we have $\displaystyle \frac{\sqrt{a^3+b^3}}{a^2+b^2}\geq \frac{1}{\sqrt{a+b}}$.

Moreover, multiply by $\displaystyle c$. We, thus, acquire

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

So,we have proved that

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

We will now apply Holder’s Inequality, that is

$\displaystyle \left(\frac{c}{\sqrt{a+b}}+\frac{b}{\sqrt{c+a}}+\frac{a}{\sqrt{b+c}}\right)^2\cdot\left[c(a+b)+b(c+a)+a(b+c)\right]\geq \left(a+b+c\right)^3$, or $\displaystyle \left(LHS\right)^2\geq \frac{(a+b+c)^3}{2(ab+bc+ca)}$.

Rewrite the sum $\displaystyle (a+b+c)^3$ as $\displaystyle (a+b+c)^2\cdot(a+b+c)$.

Then we get that:

\displaystyle \begin{aligned}\frac{(a+b+c)^3}{2(ab+bc+ca)}=\frac{(a+b+c)^2\cdot(a+b+c)}{2(ab+bc+ca)}&\geq \frac{3(ab+bc+ca)(a+b+c)}{2(ab+bc+ca)}\\&=\frac{3(a+b+c)}{2}\end{aligned}.

And finally, from the AM-GM inequality we have

$\displaystyle \frac{ 3(a+b+c)}{2}\geq \frac{3\cdot 3\sqrt[3]{abc}}{2}=\frac{9}{2}$

So, we have proved that

$\displaystyle \left(LHS\right)^2\geq \frac{9}{2}\Longrightarrow LHS\geq \frac{3}{\sqrt{2}}\Longrightarrow\frac{c\sqrt{a^3+b^3}}{a^2+b^2}+\frac{a\sqrt{b^3+c^3}}{b^2+c^2}+\frac{b\sqrt{c^3+a^3}}{c^2+a^2}\geq \frac{3}{\sqrt{2}}$, Q.E.D.