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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.

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