# Find a lower bound for a function with a cosine.

I have a function $$F(z_1,z_2) := z_1^2\cos^2(z_2)+\frac{3}{4}z_2^2$$.

I need to show that $$\forall \alpha > 0\, \exists \gamma > 0\, \forall z\, \|z\| \geq \alpha \implies F(z_1,z_2) \geq \gamma$$.

Aside from using Lagrange multipliers for this function $$F$$ with restriction $$z_1^2 + z_2^2 \geq \alpha^2$$ that give a basically unsolvable (?) equation over $$z_2$$: $$-(\alpha^2 - z_2^2)\sin(2z_2) + \frac{3}{2}z_2 -z_2\cos^2(z_2) = 0$$ do you guys have any other ideas I could use here?

I'm asking this because this is an example taken from a book so I feel like it should not be that hard, maybe some classic estimates or inequalities would work here, but I don't know many. I have tried Cauchy–Schwarz but that's basically the extents of my knowledge.

Case 1. $$z_2 <\pi/4$$: Use the identity $$\cos^2(x) = \frac12 (1 + \cos(2x))$$. This allows to write $$F(z_1,z_2) = z_1^2\cos^2(z_2)+\frac{3}{4}z_2^2 = \frac12 z_1^2 + \frac12 z_1^2 \cos(2 z_2)+\frac{3}{4}z_2^2\\ \ge \frac12 z_1^2 + \frac{1}{2}z_2^2 = \frac12 \|z\|^2 \ge \frac12 \alpha^2$$ Case 2. $$z_2 \ge \pi/4$$: $$F(z_1,z_2) = z_1^2\cos^2(z_2)+\frac{3}{4}z_2^2 \ge \frac{3}{4}z_2^2 \ge \frac{3 \pi^2}{64}$$
Clearly, if $$\alpha < \pi/4$$, then case 2 never arises.
If no notion of $$z_2$$ is available, the required bound is $$F(z_1,z_2) \ge \gamma = \min \{ \frac12 \alpha^2, \frac{3 \pi^2}{64} \}$$.