I've got this inequality, and don't know how to attack it. I assumed the converse and saw the left side can't be equal to $\pi/2$. But I feel there is something good I can't see. Any help? Thanks.
$$\text{arccosh}(a/x)+\text{arcsinh}(x/a)>\pi/2, ~~~(0<x<\pi/2,a\geq 3)$$
$$f(x)=\text{arccosh}(a/x)+\text{arcsinh}(x/a)$$ $$f{\space'}(x)=\frac{-a}{x\sqrt{a^2-x^2}}+\frac{1}{\sqrt{a^2+x^2}}\lt0$$ Hence the function $f$ is decreasing in the given interval $0<x<\dfrac{\pi}{2}$.
Take now $f\left(\dfrac{\pi}{2}\right)\approx2.852688\gt\dfrac{\pi}{2}\approx1.570779$.
Consequently, because of $f$ is decreasing, $$f(x)\gt\dfrac{\pi}{2}\text{ in the interval}\space 0\lt x\lt \dfrac{\pi}{2}$$