# An Inequality with Cubic Roots

### Solution 1

Rewrite the inequality as

$\displaystyle \sqrt[3]{4}+\sqrt[3]{9}+\sqrt[3]{25}\gt\sqrt[3]{6}+\sqrt[3]{10}+\sqrt[3]{15}.$

If $a=\sqrt[3]{2},$ $b=\sqrt[3]{3},$ $c=\sqrt[3]{5}$ then the inequality to prove becomes

$a^2+b^2+c^2\gt ab+bc+ca.$

We have several proofs for $a^2+b^2+c^2\ge ab+bc+ca,$ with the equality when $a=b=c.$

### Solution 2

For $n\ge 2,$ set $a=\sqrt[n]{5},$ $b=\sqrt[n]{3},$ $c=\sqrt[n]{2}.$ Then the inequality

$\displaystyle \sqrt[n]{4}+\sqrt[n]{9}+\sqrt[n]{25}\gt\sqrt[n]{6}+\sqrt[n]{10}+\sqrt[n]{15}$

can be rewritten as $a^2+b^2+c^2\gt ab+bc+ca,$ with a solution as above.

### Solution 3

Set

$A=\sqrt[3]{4}-\sqrt[3]{10}+\sqrt[3]{25}=2^{\frac{2}{3}}+5^{\frac{1}{3}}\left(5^{\frac{1}{3}}-2^{\frac{1}{3}}\right),\\ B=\sqrt[3]{6}-\sqrt[3]{9}+\sqrt[3]{15}=2^{\frac{1}{3}}+3^{\frac{1}{3}}+3^{\frac{1}{3}}\left(5^{\frac{1}{3}}-2^{\frac{1}{3}}\right).$

Then

\displaystyle \begin{align} A-B\;&=-2^{\frac{1}{3}}\left(3^{\frac{1}{3}}-2^{\frac{1}{3}}\right)+5^{\frac{1}{3}}\left(5^{\frac{1}{3}}-2^{\frac{1}{3}}\right)-3^{\frac{1}{3}}\left(5^{\frac{1}{3}}-3^{\frac{1}{3}}\right)\\ &\,\gt -2^{\frac{1}{3}}\left(3^{\frac{1}{3}}-2^{\frac{1}{3}}\right)+3^{\frac{1}{3}}\left(5^{\frac{1}{3}}-2^{\frac{1}{3}}\right)-3^{\frac{1}{3}}\left(5^{\frac{1}{3}}-3^{\frac{1}{3}}\right)\\ &=-2^{\frac{1}{3}}\left(3^{\frac{1}{3}}-2^{\frac{1}{3}}\right)+3^{\frac{1}{3}}\left(3^{\frac{1}{3}}-2^{\frac{1}{3}}\right)\\ &=\left(3^{\frac{1}{3}}-2^{\frac{1}{3}}\right)^2\gt 0. \end{align}

### Acknowledgment

This is a problem from R. Honsberger's Mathematical Chestnuts from Around the World (MAA, 2001, p 136.) The problem was originally offered at the 1984 Leningrad Mathematical Olympiad.

Solution 2 is by Michel Charbonneau; Solution 3 is by Daniele Pinna.