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# Inequality 43(Vo Quoc Ba Can & George Basdekis)

Problem:

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

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

Solution:

Without loss of generality, assume that $\displaystyle a\geq b\geq c$.

Since $\displaystyle a^{\frac{2}{3}}\geq b^{\frac{2}{3}}\geq c^{\frac{2}{3}}$

and

$\displaystyle \frac{1}{\sqrt{b+c}}\geq\frac{1}{\sqrt{c+a}}\geq \frac{1}{\sqrt{a+b}}$,

using Chebyshev’s Inequality we get:

$\displaystyle \frac{a^{\frac{2}{3}}}{\sqrt{b+c}}+\frac{b^{\frac{2}{3}}}{\sqrt{c+a}}+\frac{c^{\frac{2}{3}}}{\sqrt{a+b}}\geq \frac{1}{3}\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}+c^{\frac{2}{3}}\right)\cdot\left(\frac{1}{\sqrt{b+c}}+\frac{1}{\sqrt{c+a}}+\frac{1}{\sqrt{a+b}}\right)$.

Moreover, From the AM-GM Inequality we have that

$\displaystyle \frac{1}{3}\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}+c^{\frac{2}{3}}\right)\cdot\left(\frac{1}{\sqrt{b+c}}+\frac{1}{\sqrt{c+a}}+\frac{1}{\sqrt{a+b}}\right)\geq \frac{a^{\frac{2}{3}}+b^{\frac{2}{3}}+c^{\frac{2}{3}}}{\sqrt[6]{(a+b)(b+c)(c+a)}}$.

Therefore, it suffices to prove that

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

Set $\displaystyle a=x^3,b=y^3,c=z^3$.

The above inequality can be written now as

$\displaystyle 2(x^2+y^2+z^2)^2\geq 9\sqrt[3]{(x^3+y^3)(y^3+z^3)(z^3+x^3)}$, or $\displaystyle 2(x^2+y^2+z^2)^2\geq 9\sqrt[3]{xyz(x^3+y^3)(y^3+z^3)(z^3+x^3)}$.

From the AM-GM Inequality once again, we acquire that

$\displaystyle \sqrt[3]{xyz(x^3+y^3)(y^3+z^3)(z^3+x^3)}\leq \frac{x(y^3+z^3)+y(z^3+x^3)+z(x^3+y^3)}{3}$.

And thus, it is enough to check that

$\displaystyle 2(x^2+y^2+z^2)^{2}\geq 3\left[x(y^3+z^3)+y(z^3+x^3)+z(x^3+y^3)\right]$,

which is equivalent to the obvious inequality

$\displaystyle (x^2-xy+y^2)(x-y)^2+(y^2-yz+z^2)((y-z)^2+(z^2-zx+x^2)(z-x)^2\geq 0$, Q.E.D.