Prove $\int_0^{\pi/2}\frac{(a-\gamma)\sin^2\theta+1}{\sqrt{1+a\sin^2\theta}}d\theta\int_0^{\pi/2}\frac{\sin^2\theta}{\sqrt{1+a\sin^2\theta}}d\theta>1$ How to prove that
$$\int_0^{\pi/2}\frac{(a-\gamma)\cdot\sin^2\theta+1}{\sqrt{1+a\cdot \sin^2\theta}}d\theta\cdot\int_0^{\pi/2}\frac{\sin^2\theta}{\sqrt{1+a\cdot\sin^2\theta}}d\theta>1,\quad \forall a>0,$$
where $\gamma = 2-\frac{16}{\pi^2}$? Any ideas on how to prove this inequality? I found that the LHS is first increasing and then decreasing, and equality holds when $a=0$ and $a\to\infty$. But differentiating might not be a good idea.

The following inequalities can be proved through the techniques outlined in my notes (pages 134-137), i.e. by exploiting strong convexity and second-order differential equations: $$\forall a>0,\qquad \int_{0}^{\pi/2}\frac{\sin^2\theta}{\sqrt{1+a\sin^2\theta}}\,d\theta \geq \frac{\pi}{2\sqrt{4+3a}}$$
$$\forall a>0,\qquad \int_{0}^{\pi/2}\frac{1}{\sqrt{1+a\sin^2\theta}}\,d\theta =\frac{\pi}{2\,\text{AGM}(1,\sqrt{a+1})}\geq \frac{\pi}{1+\sqrt{a+1}}$$
They are enough to prove the claim in a right neighbourhood of the origin, which is a good starting point, at the very least.
