User:Klenth/Math

Given $$ \mu(\Omega) = 1 \,$$ and $$ h : \Omega \rightarrow [0, \infty]$$ being measurable, show that

$$ \sqrt{1 + A^2} \leq \int_\Omega \sqrt{1 + h^2} \; \mathrm{d}\mu $$

where $$ A = \int_\Omega h \; \mathrm{d}\mu. \,$$

Note that $$ \varphi(x) = \sqrt{1 + x^2} \,$$ is convex; then by Jensen's inequality,

$$ \varphi\left(\int_\Omega h \; \mathrm{d}\mu\right) \leq \int_\Omega \varphi(h) \; \mathrm{d} \mu$$, meaning

$$ \sqrt{1 + \left[\int_\Omega h \; \mathrm{d}\mu \right]^2 } \leq \int_\Omega \sqrt{1 + h^2} \; \mathrm{d}\mu. \,$$