Proving $a^4+2a^3 b+2ab^3+b^4 ≥ 6a^2b^2$ Prove that for any positive real numbers $a$ and $b$,$$a^4 + 2a^3b + 2ab^3 + b^4 ≥ 6a^2b^2.$$I tried using Vieta's formula to show the product of the LHS is greater than the RHS, but I don't think I am correct.

Rewrite the inequality and employ AM-GM:
$$a^4+a^3b+a^3b+ab^3+ab^3+b^4 \ge 6\sqrt[6]{a^{12}b^{12}} =6a^2b^2$$
