<s>[INST] Lower bound for $I(a,b)=\int_0^\infty \frac{dx}{\sqrt{a^2+x^2}\sqrt{b^2+x^2}} \label{elliptic integral 2}$ I am studiyng this paper which states that the integral
$$I(a,b)=\int_0^\infty \frac{dx}{\sqrt{a^2+x^2}\sqrt{b^2+x^2}}  \tag{1}$$
satisfies the inequality
\begin{align}
 \frac{\pi}{2a} \leq   I(a,b) \leq \frac{\pi}{2b} \qquad \text{for} \,\, a \ge b \tag{2}
\end{align}
The right hand side of $(2)$ is easy to establish using Cauchy-Schwarz inequality
\begin{equation}
\left(\int_{a}^{b} f(x) g(x) \mathrm{d} x\right)^{2} \leq \int_{a}^{b}(f(x))^{2} \mathrm{~d} x \int_{a}^{b}(g(x))^{2} \mathrm{~d} x \nonumber
\end{equation}

\begin{align}
  \int_0^\infty \frac{dx}{\sqrt{a^2+x^2}\sqrt{b^2+x^2}} &\leq \left(\int_0^\infty \left(\frac{1}{\sqrt{a^2+x^2}}\right)^2\,dx\right)^{1/2}\left(\int_0^\infty \left(\frac{1}{\sqrt{b^2+x^2}}\right)^2\,dx\right)^{1/2}\\
&=\left(\int_0^\infty \frac{1}{a^2+x^2}\,dx\right)^{1/2}\left(\int_0^\infty \frac{1}{b^2+x^2}\,dx\right)^{1/2}\\
&=\left(\frac{\pi}{2a}\right)^{1/2}\left(\frac{\pi}{2b}\right)^{1/2}\\
&=\frac{\pi}{2\sqrt{ab}}\\
&\leq \frac{\pi}{2\sqrt{b \cdot b}}\\
&= \frac{\pi}{2b} \qquad (\text{since} \,\ b\leq a)
\end{align}
Now, to prove the L.H.S. of $(2)$ I am unsure on how to proceed.

Edit
Following the hint below we obtain:
\begin{gather*}
    \int_0^\infty\frac{dx}{a^2+x^2} \leq \int_0^\infty\frac{dx}{\sqrt{a^2+x^2}\sqrt{b^2+x^2}}  \leq \int_0^\infty \frac{dx}{b^2+x^2}\\
     \frac{1}{a} \int_0^\infty\frac{dx}{1+x^2} \leq \int_0^\infty\frac{dx}{\sqrt{a^2+x^2}\sqrt{b^2+x^2}}  \leq \frac{1}{b}\int_0^\infty \frac{dx}{1+x^2}\\
      \frac{1}{a} \arctan(x)\Big|_0^\infty \leq \int_0^\infty\frac{dx}{\sqrt{a^2+x^2}\sqrt{b^2+x^2}}  \leq \frac{1}{b}\arctan(x)\Big|_0^\infty\\
      \frac{\pi}{2a} \leq  \int_0^\infty\frac{dx}{\sqrt{a^2+x^2}\sqrt{b^2+x^2}}  \leq \frac{\pi}{2b}
\end{gather*}
 [/INST] hint We have
$$\frac{1}{\sqrt{b^2+x^2}\sqrt{b^2+x^2}} \ge \frac1{\sqrt{a^2+x^2}\sqrt{b^2+x^2}} 
\ge \frac{1}{\sqrt{a^2+x^2}\sqrt{a^2+x^2}} .$$
Therefore...
</s>