Prove this following integral inequality Prove that $$\int_{1}^{\infty}\frac{(\sqrt{u+b})du}{u^2 + b} > \frac{\pi}{2}, b > 0$$
I know it is necessary to show the work done, but i already tried by parts and by substitution, and it does not help in nothing. My guess is that it will involves trigonometric functions, but i don't know how to get rid of the b in the integrals. Any tip?
 A: Here's a partial answer / an idea for an approach.

For $b \geq 0$, let
$$\ell(b) = \int_1^\infty \frac{\sqrt{u+b}}{u^2+b} \; \mathrm{d}u$$
We see that
$$\ell'(b) = \int_1^\infty \frac{\partial}{\partial b} \frac{\sqrt{u+b}}{u^2+b} \; \mathrm{d}u = \int_1^\infty \frac{u^2 - 2u - b}{2\sqrt{b+u}(b+u^2)^2} \; \mathrm{d}u \geq \int_1^\infty \frac{u^2 -2u - b}{2(b+u^2)^{5/2}}\; \mathrm{d}u,$$
and we can actually do this last integral! It turns out to be
$$\int \frac{u^2 - 2u - b}{2(b+u^2)^{5/2}} \; \mathrm{d}u = \frac{b (2-3 u)-u^3}{6 b \left(b+u^2\right)^{3/2}} + c,$$
so
$$\ell'(b) \geq \int_1^\infty \frac{u^2 - 2u - b}{2(b+u^2)^{5/2}} \; \mathrm{d}u = \frac{\frac{1}{\sqrt{b+1}}-1}{6 b}.$$
Now, for any $b > 0$, we have
$$\ell(b) = \ell(0) + \int_0^b \ell'(t) \; \mathrm{d}t \geq \ell(0) + \int_0^b \frac{\frac{1}{\sqrt{t+1}}-1}{6 t} \; \mathrm{d}t = 2 + \frac{1}{3} \log \left(\frac{2}{1 + \sqrt{1 + b}}\right).$$
For $b \leq 38$ this is enough to show $\ell(b) > \pi/2$. With a better bound on $\ell'$ we might be able to improve this.
