# 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.)

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- An Extension of the AM-GM Inequality
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- Rearrangement Inequality
- Chebyshev Inequality
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- Dorin Marghidanu's Inequality in Many Variables
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- An Inequality with Integrals and Radicals
- Twin Inequalities in Four Variables: Twin 1
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- Simple Inequality with a Variety of Solutions
- A Partly Cyclic Inequality in Four Variables
- Dan Sitaru's Inequality by Induction
- An Inequality in Three (Or Is It Two) Variables
- An Inequality in Four Weighted Variables
- An Inequality in Fractions with Absolute Values
- Inequalities with Double And Triple Integrals
- An Old Inequality
- Dan Sitaru's Amazing, Never Ending Inequality
- Leo Giugiuc's Exercise
- Another Inequality with Logarithms, But Not Really
- A Cyclic Inequality of Degree Four
- An Inequality Solved by Changing Appearances
- Distances to Three Points on a Circle
- An Inequality with Powers And Logarithm
- Four Integrals in One Inequality
- Same Integral, Three Intervals
- Dorin Marghidanu's Inequality with Generalization
- Dan Sitaru's Inequality with Three Related Integrals and Derivatives
- An Inequality in Two Or More Variables
- An Inequality in Two Or More Variables II
- A Not Quite Cyclic Inequality
- Dan Sitaru's Inequality: From Three Variables to Many in Two Ways
- An Inequality with Sines But Not in a Triangle
- An Inequality with Angles and Integers
- Sladjan Stankovik's Inequality In Four Variables
- An Inequality with Two Pairs of Triplets
- A Refinement of Turkevich's Inequality
- Dan Sitaru's Exercise with Pi and Ln
- Problem 4165 from Crux Mathematicorum

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