# Inequality from Math Phenomenon

### Statement

Leo Giugiuc has kindly posted a problem by Dan Sitaru at the CutTheKnotMath facebook page. He later posted a solution (Solution 2) and encouraged Daniel Dan to post another one (Solution 3).

There at least two different mathematical statements that go under the name of Inequality from Math Phenomenon (Pafnuty Chebyshev (1821-1894), a famous Russian mathematician.) One in the realm of Probability Theory is of fundamental importance, the other - the subject of the present page - is an extremely useful and powerful tool for solving olympiad-kind inequalities

Assume $a,b,c\gt 0$ and $n$ a positive integer. Then

$\displaystyle \frac{a^{2n+1}}{\sqrt{bc}}+\frac{b^{2n+1}}{\sqrt{ac}}+\frac{c^{2n+1}}{\sqrt{ab}}\ge a^{2n}+b^{2n}+c^{2n}.$

### Solution 1

Assume, the sequence $a,b,c$ is ordered and observe that the sequence $\displaystyle\frac{a}{\sqrt{bc}}=\frac{a^{3/2}}{\sqrt{abc}},$ ... is similarly ordered. We are thus in a position to apply Chebyshev Inequality:

$\displaystyle 3\left(\frac{a^{2n+1}}{\sqrt{bc}}+\frac{b^{2n+1}}{\sqrt{ac}}+\frac{c^{2n+1}}{\sqrt{ab}}\right)\ge \left(\frac{a}{\sqrt{bc}}+\frac{b}{\sqrt{ac}}+\frac{c}{\sqrt{ab}}\right)(a^{2n}+b^{2n}+c^{2n}).$

But, by the AM-GM inequality,

$\displaystyle \frac{1}{3}\left(\frac{a}{\sqrt{bc}}+\frac{b}{\sqrt{ac}}+\frac{c}{\sqrt{ab}}\right)\ge \sqrt[3]{\frac{a}{\sqrt{bc}}\frac{b}{\sqrt{ac}}\frac{c}{\sqrt{ab}}}=\sqrt[3]{\frac{abc}{\sqrt{a^2b^2c^2}}}=1.$

### Solution 2

The given inequality is equivalent to

$\displaystyle a^{2n+\frac{3}{2}}+b^{2n+\frac{3}{2}}+c^{2n+\frac{3}{2}}\ge \sqrt{abc}(a^{2n}+b^{2n}+c^{2n}).$

Since the functions $f(x)=x^{2n}$ and $f(x)=x^{\frac{3}{2}}$ are strictly increasing for positive $x,$ then, by Chebyshev Inequality, we get

$\displaystyle a^{2n+\frac{3}{2}}+b^{2n+\frac{3}{2}}+c^{2n+\frac{3}{2}} \ge \frac{1}{3}(a^{2n}+b^{2n}+c^{2n})\left(a^{\frac{3}{2}}+b^{\frac{3}{2}}+c^{\frac{3}{2}}\right).$

Hence, it is sufficient to prove

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

But this is equivalent to $\displaystyle\frac{1}{3}\left(a^{\frac{3}{2}}+b^{\frac{3}{2}}+c^{\frac{3}{2}}\right)\ge\sqrt{abc}.$ By the AM-GM inequality this is indeed so, and this completes the proof.

### Solution 3

Observe that, as an application of the Cauchy-Schwarz inequality

$\displaystyle\left(\sum_{i=1}^{n}x_iy_i\right)^{2}\le\sum_{i=1}^{n}x_{i}^{2}\sum_{i=1}^{n}y_{i}^{2},$

we get

$\displaystyle\left(\sum_{i=1}^{n}a_i\right)^{2}\le\sum_{i=1}^{n}b_{i}\sum_{i=1}^{n}\frac{a_{i}^{2}}{b_i},$

We modify the left-hand side of the required inequality and then apply the latter:

\displaystyle\begin{align} \frac{a^{2n+1}}{\sqrt{bc}}+\frac{b^{2n+1}}{\sqrt{ac}}+\frac{c^{2n+1}}{\sqrt{ab}}&=\frac{a^{4n}}{a^{2n-1}\sqrt{bc}}+\frac{b^{4n}}{b^{2n-1}\sqrt{ac}}+\frac{c^{4n}}{c^{2n-1}\sqrt{ab}}\\ &\ge\frac{\left(a^{2n} +b^{2n} +c^{2n}\right)^2}{a^{2n-1}\sqrt{bc}+b^{2n-1}\sqrt{ac}+c^{2n-1}\sqrt{ab}}\\ &\ge\frac{\left(a^{2n} +b^{2n} +c^{2n}\right)^2}{a^{2n} +b^{2n} +c^{2n}}\\ &=a^{2n} +b^{2n} +c^{2n}. \end{align}

(The last step folows from the Rearrangement Inequality.)