in preparation for an exam I wanted to show that for $\mu(\Omega)=1$ and $h:\Omega \rightarrow [0,\infty]$ measurable the following inequality holds:
$\sqrt{1 + (\int_{\Omega} h d\mu)^2} \leq \int_{\Omega} \sqrt{1+h^2} d\mu$
I'm having trouble dealing with the squared integral. I thought of doing something like this:
$1 + (\int_{\Omega} h d\mu)^2 = \int_{\Omega} 1 d\mu + \int_{\Omega}\int_{\Omega} h(x)\cdot h(y) dx dy$ But that's probably not allowed right?
Thanks for any help!
Hint: Apply Jensen's inequality to the convex function $x\mapsto \sqrt{1+x^2}$.
Trivia: If $\Omega=[0,1]$, $\mu=\lambda$ and $h=f'$ for some increasing $C^1$ function $f$ with $f(0)=0$ and $f(1)=1$, then the inequality reads $$ \sqrt{1^2+1^2}=\sqrt{1+\left(\int_0^1 f'(x)\,\mathrm dx\right)^2}\leq \int_0^1 \sqrt{1+f'(x)^2}\,\mathrm dx $$ meaning that the shortest path between the points $(0,0)$ and $(1,1)$ is obtained when travelling in a straight line, since the right-hand side is the arc length of the path obtained by travelling along $f$.