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?

Yes, you can!
A full expending gives
$$\sum_{cyc}x^6\sum_{cyc}(x+1)^2-\left(\sum_{cyc}(x^4+x^3)\right)^2=$$
$$=\sum_{cyc}(x^6y^2+x^6z^2-2x^4y^4+2x^6y-2x^4y^3-2x^4z^3+2x^6-2x^3y^3)=$$
$$=\sum_{cyc}x^2y^2(x^2-y^2)^2+2\sum_{cyc}xy(x+y)(x^2+xy+y^2)(x-y)^2+\sum_{cyc}(x^3-y^3)^2\geq0.$$
Actually, the inequality that you named Holder's inequality, named Cauchy-Schwarz inequality. 
