# How to prove the first inequality?

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?

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The function $\phi \left( x \right) = \sqrt{1 + x^2}$ is convex for $x > 0$. Using Jensen's inequality $$\phi \left( \mu \left( h \right) \right) \leqslant \mu \left( \phi \left( h \right) \right)$$ where $\mu \left( g \right) = \int_X g \mathrm{d} \mu$ for any $g$.