# Inequality with Powers And Radicals

### Problem

### Proof 1

The sought inequality is equivalent to

$\displaystyle\frac{1}{\sqrt[3]{abc}}\sum_{cycl}\sqrt[6]{ab^2c^3}\ge\frac{1}{\sqrt[3]{abc}}\sum_{cycl}\sqrt[30]{a^9b^{10}c^{11}}.$

which translates into

$\displaystyle\sum_{cycl}\sqrt[6]{\frac{a}{b}}\ge\sum_{cycl}\sqrt[30]{\frac{a}{b}}.$

Defining $\displaystyle x=\sqrt[30]{\frac{a}{b}},\;$ etc, the problem reduces to showing that

$\sum_{cycl}x^5\ge\sum_{cycl}x,\;$ provided $x,y,z\gt 0\;$ and $xyz=1.$

The AM-GM inequality, yields

$x^5+1+1+1+1\ge 5x,\\ y^5+1+1+1+1\ge 5y,\\ z^5+1+1+1+1\ge 5z.$

Summing up shows that

$\displaystyle\begin{align} \sum_{cycl}x^5 &\ge 5\sum_{cycl}x-12\\ &=\sum_{cycl}x+4\left(\sum_{cycl}x-3\right)\\ &=\sum_{cycl}x+4\left(\sum_{cycl}x-3\sqrt[3]{\prod_{cycl}x}\right)\\ &\ge \sum_{cycl}x \end{align}$

because, by the AM-GM inequality, $\displaystyle x+y+z\ge 3\sqrt[3]{xyz}.$ The equality is attained when $x=y=z=1,\;$ i.e., when $a=b=c.$

### Proof 2

By the AM-GM inequality,

$\displaystyle\begin{align} 7(ab^2c^3)^{\frac{1}{6}}+4(a^3bc^2)^{\frac{1}{6}}+4(a^2b^3c)^{\frac{1}{6}} &\ge (a^7b^{14}c^{21}a^{12}b^4c^8a^8b^{12}c^4)^{\frac{1}{6\times;15}}\\ &= 15(a^{27}b^{30}c^{33})^{\frac{1}{90}}\\ &= 15(a^9b^{10}c^{11})^{\frac{1}{30}}. \end{align}$

Similarly,

$4(ab^2c^3)^{\frac{1}{6}}+7(a^3bc^2)^{\frac{1}{6}}+4(a^2b^3c)^{\frac{1}{6}} \ge 15(a^{11}b^9c^{10})^{\frac{1}{30}},\\ 4(ab^2c^3)^{\frac{1}{6}}+4(a^3bc^2)^{\frac{1}{6}}+7(a^2b^3c)^{\frac{1}{6}} \ge 15(a^{10}b^{11}c^{9})^{\frac{1}{30}}.$

Adding the three and dividing by $15\;$ gives the required inequality.

### Proof 3

This is also a direct consequence of Muirhead's inequality. Indeed, let $\displaystyle\mathbf{\alpha}=\{\frac{3}{6},\frac{2}{6},\frac{1}{6}\}\;$ and $\displaystyle\mathbf{\beta}=\{\frac{11}{30},\frac{10}{30},\frac{9}{30}\}.\;$ Then $\mathbf{\alpha}\;$ *majorizes* $\mathbf{\beta}\;$ which imediately implies the given inequality.

### Acknowledgment

The problem from his book *Math Accent* has been posted by Dan Sitaru at the CutTheKnotMath facebook page. He also added two solutions: Proof 1 by Dang Thanh Tung and Proof 2 by Ravi Prakash.

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