Prove $(x^{3} + x^{4} +y^{3} + y^{4} + z^{3} + z^{4})^{2} \le (x^{6} + y^{6} + z^{6})[ (1+x)^{2} + (1+y)^{2} + (1+z)^{2} ] $ Prove 
$$ (x^{3} + x^{4} +y^{3} + y^{4} + z^{3} + z^{4})^{2} \le (x^{6} + y^{6} + z^{6})[ (1+x)^{2} + (1+y)^{2} + (1+z)^{2} ] $$
if $x,y,z \ge 0$

I made this problem using Holders inequality, notice that
$$ (1+x)x^{3} + (1+y)y^{3} + (1+z)z^{3} \le \left[ (x^{3})^{2} + (y^{3})^{2} + (z^{3})^{2} \right]^{1/2} \left[ (1+x)^{2} + (1+y)^{2} + (1+z)^{2} \right]^{1/2} $$
then squaring both sides we get 
$$ (x^{3} + x^{4} +y^{3} + y^{4} + z^{3} + z^{4})^{2} \le (x^{6} + y^{6} + z^{6})[ (1+x)^{2} + (1+y)^{2} + (1+z)^{2} ] $$
My question is, will this inequality be obvious to be solved without Holders inequality? if so then what is the alternative solution?

This is just Cauchy inequality $$(a^2+b^2+c^2)(a'^2+b'^2+c'^2)\geq (aa'+bb'+cc')^2$$
where $a= 1+x$ and $a'=x^3$...
