# Inequality $\frac{a + \sqrt{ab} + \sqrt[3]{abc}}{3} \leq \sqrt[3]{a \cdot \frac{a+b}{2} \cdot \frac{a+b+c}{3}}.$

Someone can to help me with a hint in the following problem:

Show that for any $a,b,c>0$, $$\frac{a + \sqrt{ab} + \sqrt[3]{abc}}{3} \leq \sqrt[3]{a \cdot \frac{a+b}{2} \cdot \frac{a+b+c}{3}}.$$

I have tried using the Hölder inequality, but can not apply it efficiently.

Thanks!

• Have you tried to use the inequality involving arithmetic and geometric mean? – Ulrik Jan 5 '14 at 0:53
• ^^^^This inequality looks ripe for the picking by some form of Power Mean. – Ayesha Jan 5 '14 at 0:55
• Yeah. But I don't had success with the GM-inequality.. – Walner Jan 5 '14 at 0:57
• May you include the source? – chubakueno Jan 5 '14 at 1:34
• This is a problem of a graduate exam of my university. Here is the link: docs.google.com/file/d/0B554_AgrU7nNRUY0VTBvTmNJd28/edit But is in Portuguese. Is the question 9, in the sentence C. – Walner Jan 5 '14 at 1:38

First, prove the following inequality: For $a_{ij}\in \mathbb{R}_{\geq0},1\leq i \leq m, 1\leq j \leq n$,
$\sum_{i=1}^m \sqrt[n]{\prod_{j=1}^n{a_{ij}}}\leq \prod_{j=1}^n \sqrt[n]{ \sum_{i=1}^m {a_{ij}}}$.
(by induction on $n$ fixing $m$, and using Holder's ineq with $x_i=\sqrt[n]{\prod_{j=1}^{n-1} {a_{ij}}}, y_i=\sqrt[n]{a_{in}}, p=\frac{n}{n-1}, q=n$ for induction step)
Then it is quite trivial, by letting $(a_{ij})=\left( \begin{array}{ccc} x & x & x \\ x & \sqrt{xy} & y \\ x & y & z \end{array} \right)\in \mathbb{R}^{3\times 3}$ in the above and using AM-GM ineqs.