Prove $ (x^{3} + x^{4} +y^{3} + y^{4} + z^{3} + z^{4})^{3} \le (x^{9} + y^{9} + z^{9})[ (1+x)^{3/2} + (1+y)^{3/2} + (1+z)^{3/2} ]^{2} $ Prove 
$$ (x^{3} + x^{4} +y^{3} + y^{4} + z^{3} + z^{4})^{3} \le (x^{9} + y^{9} + z^{9})[ (1+x)^{3/2} + (1+y)^{3/2} + (1+z)^{3/2} ]^{2} $$
if $x,y,z \ge 0$

I made this problem using Holders inequality, notice that
$$ (1+x)x^{3} + (1+y)y^{3} + (1+z)z^{3} \le \left[ (x^{3})^{3} + (y^{3})^{3} + (z^{3})^{3} \right]^{1/3} \left[ (1+x)^{3/2} + (1+y)^{3/2} + (1+z)^{3/2} \right]^{2/3} $$
then taking cube both sides we get 
$$ (x^{3} + x^{4} +y^{3} + y^{4} + z^{3} + z^{4})^{3} \le (x^{9} + y^{9} + z^{9})[ (1+x)^{3/2} + (1+y)^{3/2} + (1+z)^{3/2} ]^{2} $$
My question is, will this inequality be obvious to be solved without Holders inequality? if so then what is the alternative solution?

Directly citing the Hölder inequality is the shortest solution. If one wants to reduce this to some somewhat more elementary arguments, one can use the inequality of the arithmetic and cubic mean.
For non-negative $a_k$ and positive $b_k$ we get, using the weighted mean value inequalities, that is, Jensen's inequality for a discrete measure, and using weights $w_k=b_k^3$,
\begin{align}
\frac{\sum a_kb_k^2}{\sum b_k^3}
=\frac{\sum w_k\frac{a_k}{b_k}}{\sum w_k}
\le\sqrt[3]{\frac{\sum w_k\left(\frac{a_k}{b_k}\right)^3}{\sum w_k}}
=\sqrt[3]{\frac{\sum a_k^3}{\sum b_k^3}}
\\~\\
\implies \left(\sum a_kb_k^2\right)^3 \le \sum a_k^3\left(\sum b_k^3\right)^2
\end{align}
Now set $a_1=x^3$, $b_1=\sqrt{1+x}$ etc.
