Prove inequality $(x^3+2y^3+16\sqrt2)^4 \le 81(x^4+y^4+16)^3$

Question

Prove inequality $$(x^3+2y^3+16\sqrt2)^4 \le 81(x^4+y^4+16)^3$$

My work

$$(x^3+y^3+y^3+(2\sqrt2)^3)^4 \le 3^4(x^4+y^4+2^4)^3$$

Any hints?

Answer 1

Since $$|x^3+2y^3+16\sqrt2|\leq|x|^3+2|y|^3+16\sqrt2,$$ it's enough to prove our inequality for non-negative variables.

Thus, by Holder $$81(x^4+y^4+16)^3=(1+16+64)(x^4+y^4+16)^3\geq$$ $$\geq\left(\sqrt[4]{1\cdot\left(x^4\right)^3}+\sqrt[4]{16\cdot(y^4)^3}+\sqrt[4]{64\cdot16^3}\right)^4=(x^3+2y^3+16\sqrt2)^4.$$