# Prove that for all real numbers $x$ and $y$ that $x^4+y^4+8 \geq 8xy$.

Prove that for all real numbers $x$ and $y$ that $x^4+y^4+8 \geq 8xy$.

It looks like I could use AM-GM on the $xy$ term, but then I would get a term on the RHS greater than what is already on the RHS. I don't know if we could still prove it then.

• The really scary part is that $x^4+y^4\color{red}-8=8xy$ is so incredibly close to an ellipse. – Lucian Dec 27 '15 at 17:31

By AM-GM (which we can use because $x^4,y^4\ge 0$): $$x^4+y^4+4+4\ge 4\sqrt[4]{x^4\cdot y^4\cdot 4\cdot 4}=8|xy|\ge 8xy$$
With equality iff $x^4=y^4=4$ and $xy\ge 0$, i.e. iff $x=y=\pm \sqrt{2}$.
We have: $x^4+y^4 +8\geq 2x^2y^2+8=2(x^2y^2+4)=2((xy)^2+2^2)\geq 2(2(xy)\cdot 2)=8xy$