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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.

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You can do this with two cases.

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 $F(z_1,z_2) \ge \gamma = \min \{ \frac12 \alpha^2, \frac{3 \pi^2}{64} \}$.

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