I've been learning about $L^p$-spaces and came across this problem in Rudin's Real and Complex Analysis:
Suppose $\mu(\Omega)=1$ and $h:\Omega \to [0,\infty]$ is measurable. If $A= \int_\Omega {h d\mu}$, prove that: $$ \sqrt{1+A^2} \leq \int_\Omega \sqrt{1+h^2}d\mu \leq 1+A $$
I've already shown that $\sqrt{1+A^2} \leq \int_\Omega \sqrt{1+h^2}d\mu$ using Jensen's inequality applied to the convex function $\varphi(x)=\sqrt{1+x^2}$. I can't seem to show the latter inequality. Any help or hints would be appreciated. Thank you in advance!
It is simpler than you are expecting. Since $h \ge 0$ you have $1 + h^2 \le 1 + 2h + h^2 = (1+h)^2$ so that $\sqrt{1 + h^2} \le 1+h$. Now integrate this over $\Omega$.