Assume $\mu(X)=1$ and $h\ge 0$ is measurable, if $A=\int_{X}hd\mu$, prove that
$\sqrt{1+A^{2}}\le \int_{X}\sqrt{1+h^{2}}d\mu\le 1+A$
I am wondering how to prove the first part of the inequality. It is straightforward that $\sqrt{1+h^{2}}\le 1+h$, and hence the second inequality. But I am at lost how to prove the first identity.
An elementary approach is to expand $\sqrt{1+h^{2}}$ and compare both sides. The first few terms are $1+\frac{1}{2}h^{2}+O(h^{4})$. So integrating it we get $1+\frac{1}{2}\int h^{2}+\int O(h^{4})$. And squaring on both sides would give $1+\frac{1}{4}\left(\int h^{2}\right)^{2}+\left(\int O(h^{4})\right)^{2}+\int h^{2}+2\int O(h^{4})+\int h^{2}\int O(h^{4})$
The problem is the $O(h^{4})$ term is not really bounded and can be negative. So the above expansion does not help to prove the inequality. I am wondering if there is some easier ways to attack this problem. Rudin suggest to use a function $g\in C_{c}(X)$ to approximate $h$, and the inequality has a geometric meaning in terms of $g$. But what is it?