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} ] $

Question

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 $x,y,z \ge 0$

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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?

Answer 1

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 $a'=x^3$...

Answer 2

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.