# Proving an inequality in 3 dimensional standard Lebesgue measure and geometrical interpretation

Suppose $\Omega=[0,1]\times [0,1]\times[0,1]\subset \mathbb{R}^3$ and $q:\Omega\to [0,\infty]$ is measurable. If $B:=\int_\Omega q d\mu$, prove that $$\sqrt{1+B^2}\le \int_\Omega\sqrt{1+q^2}d\mu\le 1+B$$

where $\mu$ stands for the standard 3-dimensional Lebesgue measure. Give a geometric interpretation of these inequalities.

Hi I was trying to understand how to tackle this problem, but I couldn't even start. Any hints would be greatly appreciated. Thank you very very much.

The key lies in the simple inequality $$\sqrt{1+q^2} \leqslant 1+q,$$ which is an easy consequence of squaring both sides: $2q \geqslant 0 \implies (1+q)^2 \geqslant 1+q^2$. Using this on the integral immediately gives $$\int_{\Omega} \sqrt{1+q^2} \, d\mu \leqslant \int_{\Omega} (1+q) \, d\mu = 1+B.$$ The other inequality we can find by looking at $$\left( \int_{\Omega} \sqrt{1+q^2} \, d\mu \right)^2 - \left( \int_{\Omega} q \, d\mu \right)^2 = \left( \int_{\Omega} \left(\sqrt{1+q^2}-q\right) d\mu \right) \left( \int_{\Omega} \left( \sqrt{1+q^2}+q \right) d\mu \right).$$ Both integrands are bounded below by $1$, the latter obviously, the former by the first inequality we mentioned. Hence the whole lot is bounded below by $1$, and the left inequality follows.