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!
 A: 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.
A: 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 $a=b=c$
