Prove that for any real numbers $a, b, a^4+b^4+2 \geqslant 4ab$.

Question

I was given a hint to use AM-GM here, but I am no sure how to create an arithmetic or geometric mean here because of the exponents.

Answer 1

As the hint should be an answer, I post it as community wiki.

Use AM-GM on $a^4$, $b^4$, $1$ and $1$.

\begin{align} \frac{a^4+b^4+1+1}{4} & \ge\sqrt[4]{a^4\cdot b^4 \cdot 1 \cdot 1} \\ a^2+b^2+2 &\ge 4|ab| \ge 4ab \end{align}

Answer 2

I've got another way of solving it, one without AM-GM:

\begin{align} a^4+b^4+2 & \ge 4ab \\ a^4-2a^2+1+b^4-2b^2+1+2a^2+2b^2 & \ge 4ab \\ (a^2-1)^2+(b^2-1)^2+2a^2-4ab+2b^2 & \ge 0 \\ (a^2-1)^2+(b^2-1)^2+2(a-b)^2 & \ge 0 \\ \end{align}

The sum of squares is always greater than $0$, because the square of any number is greater than or equal to $0$. Hence the inequality $a^4+b^4+2 \ge 4ab$ holds true.

Answer 3

Two times AM-GM: $$ (a^4+b^4)+2 \geq 2a^2b^2+2=2(a^2b^2+1) \geq 4 ab $$