Inequality with angles of a triangle Let $\alpha$, $\beta$ and $\gamma$ be the three angles in a non-right triangle. How can i prove the following inequality?
$\dfrac{1+\sin^2(\alpha)}{\cos^2(\alpha)}+\dfrac{1+\sin^2(\beta)}{\cos^2(\beta)}+\dfrac{1+\sin^2(\gamma)}{\cos^2(\gamma)}\ge \dfrac{1+\sin(\alpha)\sin(\beta)}{1-\sin(\alpha)\sin(\beta)}+\dfrac{1+\sin(\beta)\sin(\gamma)}{1-\sin(\beta)\sin(\gamma)}+\dfrac{1+\sin(\alpha)\sin(\gamma)}{1-\sin(\alpha)\sin(\gamma)}$

$$\sum_{cyc}\frac{(2bc)^2+16\Delta^2}{(b^2+c^2-a^2)^2}\geq \sum_{cyc}\frac{4 R^2+ab}{4R^2-ab}$$
is equivalent to
$$ 16\Delta^2\sum_{cyc}\frac{1}{(b^2+c^2-a^2)^2} \geq \sum_{cyc}\frac{ab}{4R^2-ab} $$
or to
$$\sum_{cyc}\frac{1}{(b^2+c^2-a^2)^2} \geq \sum_{cyc}\frac{1}{(b^2+c^2-a^2)^2+4(a-b)bc^2}$$
which can be proved by bashing. Although I suspect there are more efficient ways, based on the convexity of $f(x)=\frac{1+x}{1-x}$ over $(-1,1)$.
