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.

You can easily prove it using AM-GM Inequality or Lagrange method. Using AM-GM Inequality,
$$ \frac {a^4 + b^4}{2} \ge a^2.b^2 $$
$$ \frac {a^2 + b^2}{2} \ge a.b$$
$$ \text {Also, you can write }a^3.b+a.b^3 = ab.(a^2+b^2)$$
$$ \text {So, } a^4 + b^4 +2a^3.b+2a.b^3 \ge 6.a^2.b^2$$
