Proving $f(x,y)=\cos^4y-(1-\tan^4x-\sin^2y)(\sin^2x\sin^2y+\cos^2y)^2\geq0$ in some open neighborhood of $(0,0)$ Time is ticking and I must prove the Hourglass (or Figure 8) Inequality, which states 

$f(x,y)\geq 0$ in some open neighborhood of $(0,0)$, where 
  $$f(x,y)=\cos^4 y -(1-\tan^4 x - \sin^2 y)(\sin^2 x \sin^2 y + \cos^2 y)^2$$

This inequality implies an interesting continuity result.
The reason I named it the Figure 8 Inequality is because of the level curves you see for $f(x,y)=0.002$:

Unfortunately, it appears that the circuit is closing in on $(0,0),$ so there is absolutely no room for error. The inequality must be proven in a very precise manner, with each step of the form $A \ge B$ being carried so that $A-B$ is extremely small. I do not know how you could possibly make each leap to the conclusion razor sharp.

Let us prove that $f(x, y) \ge 0$ for $|x| < \frac{1}{2}$ and $|y| < \frac{\pi}{6}$.
When $|x| < \frac{1}{2}$ and $|y| < \frac{\pi}{6}$, we have $|\sin x| \le |x| \le |\tan x|$,
$\frac{3}{4} \le \cos^2 y \le 1$, and $1-\tan^4 x - \sin^2 y > 0$.
Denote $v = \cos^2 y \in [\frac{3}{4}, 1]$. We have
\begin{align}
f(x, y) &\ge \cos^4 y - (1 - x^4 - \sin^2 y)(x^2 \sin ^2 y + \cos^2 y)^2\\
&= v^2 - (1 - x^4 - (1-v))(x^2 (1-v) + v)^2 \\
&= -(1-x^2)^2 v^3+ (x^4-x^2+1)^2 v^2+(-2 x^8+2 x^6-x^4) v+x^8\\
&\ge -(1-x^2)^2 v^2 + (x^4-x^2+1)^2 v^2+(-2 x^8+2 x^6-x^4) v+x^8\\
&= (x^8-2 x^6+2 x^4) v^2+(-2 x^8+2 x^6-x^4) v+x^8\\
&\ge (x^8-2 x^6+2 x^4) v\cdot \frac{3}{4} +(-2 x^8+2 x^6-x^4) v+x^8\\
&= (-x^8 + 2x^6 + 2x^4)v + 4x^8(1-v)\\
&\ge 0.
\end{align}
We are done.
