Using calculus to prove an algebraic inequality Let $$f(x,y):=4\pi \left(\frac{y}{\tan(\frac{\pi}{2}x)}+\frac{x}{\tan(\frac{\pi}{2}y)}\right)+16(1+x+y)-\frac{16}{3}\frac{(x+y+xy)^2}{xy}. $$
Prove that $f(x,y) \ge 0$, for $0\le x <1, 0\le y<1$.
The figure of the function from WolframAlpha indicates the above claim is true. I'm wondering how to show this by hand?

Using $\frac{2}{\pi}u \le \sin u \le u$
for all $u\in [0, \pi/2]$, we have
$$\frac{1}{\tan(\pi x/2)}
= \frac{\sin(\pi/2 - \pi x/2)}{\sin(\pi x/2)} \ge \frac{\frac{2}{\pi}\cdot (\pi/2 - \pi x/2) }{\pi x/2} = \frac{2}{\pi}\cdot \frac{1 - x}{x}.$$
Then, it suffices to prove that
$$4\pi \cdot \left(y\cdot \frac{2}{\pi}\cdot \frac{1 - x}{x} + x\cdot \frac{2}{\pi}\cdot \frac{1 - y}{y}\right) + 16(1 + x + y) - \frac{16}{3}(x + y + xy)^2/(xy) \ge 0$$
or
$$\frac{8(x + y + xy)(x + y - 2xy)}{3xy} \ge 0$$
which is clearly true.
We are done.
