I would like to prove that the following function $f :\mathbb{R}^2\to\mathbb{R}$ has a global minimum:
$f(x,y)=(x-2y)^4+64xy$
I've found the gradient and the hessian of this function, along with it's local minima. I need to prove that those local minima are also global minimums.
$f$ has strict local minima at $f(1,-1/2)=-16$ and $f(-1,1/2)=-16$
I think that what I need to show is that $-16$ is a lower bound of this function, and then conclude that's a global minimum, but I didn't manage to do so.
Please advise.
Thank you.
Best Answer
Put $u = x, v = -2y \implies f(u,v) = (u+v)^4 - 32uv$. Thus you prove: $f(u,v) \ge -16 \iff (u+v)^4 - 32uv + 16 \ge 0$. If $u, v \ge 0$, then by AM-GM inequality: $(u+v)^4 \ge \left(2\sqrt{uv}\right)^4 = 16(uv)^2\implies (u+v)^4-32uv+16 \ge 16(uv)^2 - 32uv+16 = 16(uv-1)^2\ge 0$. Equality occurs when $uv = 1$ which corresponds to $(x,y) = (\pm 1, \mp \frac{1}{2})$ that was discussed before. If $u \le 0, v \ge 0$, then this corresponds to $x \le 0, y \le 0$, thus $xy \ge 0$, and $f(x,y) = (x-2y)^4 + 64xy \ge 64xy \ge 0 > -16$. If $u \ge 0, v \le 0$, then $x \ge 0, y \ge 0\implies f(x,y) \ge 64xy \ge 0 > -16$. If $u \le 0, v \le 0$, then put $m = -u, n = -v \implies f(m,n) = (-m-n)^4 - 32(-m)(-n)= (m+n)^4-32mn\ge -16$ by AM-GM again for the pair $(m,n)$ with $m \ge 0, n\ge 0$. Thus we've shown $f(x,y) \ge -16$ for all $(x,y) \in \mathbb{R^2}$, confirming $-16$ is the global minimum for $f$.