# 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? Jan 5 '14 at 0:53
• ^^^^This inequality looks ripe for the picking by some form of Power Mean. Jan 5 '14 at 0:55
• Yeah. But I don't had success with the GM-inequality.. Jan 5 '14 at 0:57
• May you include the source? 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. 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.

• And I think I saw this ineq in some olympiad problem set. Jan 5 '14 at 2:22
• Indeed, this solves! Thanks! I don't known this inequality! Again, thanks! Jan 5 '14 at 14:18

By AM-GM inequality: $$\sqrt{ab}\le\sqrt[3]{ab.\frac{a+b}{2}}$$

We prove that: $$\sqrt[3]{\frac{2}{a+b}.\frac{3}{a+b+c}}\left(a+\sqrt[3]{ab.\frac{a+b}{2}}+\sqrt[3]{abc}\right)\le3\sqrt[3]{a}$$

Also by AM-GM: $$\sqrt[3]{\frac{2a}{a+b}.\frac{3a}{a+b+c}}\le\frac{1+\frac{2a}{a+b}+\frac{3a}{a+b+c}}{3}$$ $$\sqrt[3]{\frac{3b}{a+b+c}}\le\frac{2+\frac{3b}{a+b+c}}{3}$$

$$\sqrt[3]{\frac{2b}{a+b}.\frac{3c}{a+b+c}}\le\frac{1+\frac{2b}{a+b}+\frac{3c}{a+b+c}}{3}$$ Sum up term by term, we obtain complete proof!

Equality holds iff $$a=b=c$$