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!

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 $a=b=c$
