TITLE: Show that $\left(\int_{0}^{1}\sqrt{f(x)^2+g(x)^2}\ dx\right)^2 \geq \left(\int_{0}^{1} f(x)\ dx\right)^2 + \left(\int_{0}^{1} g(x)\ dx\right)^2$
QUESTION [5 upvotes]: Show that
  $$
\left( \int_{0}^{1} \sqrt{f(x)^2+g(x)^2}\ \text{d}x \right)^2
\geq
\left( \int_{0}^{1} f(x)\ \text{d}x\right)^2
+ \left( \int_{0}^{1} g(x)\ \text{d}x \right)^2
$$
  where $f$ and $g$ are integrable functions on $\mathbb{R}$.

That inequality is a particular case. I want to approximate the integral curves using some inequalities who imply this inequality.

REPLY [2 votes]: Suppose that $f$ and $g$ are continuous functions. Define
$$\phi(t) = \left(\int_0^t \sqrt{f(s)^2 + g(s)^2} ds\right)^2 -\left(\int_0^t f(s) ds\right)^2 - \left(\int_0^t g(s) ds\right)^2.$$
It is obvious that $\phi(0) =0$ and
$$\phi'(t) = 2\left[\int_0^t \sqrt{f(t)^2 + g(t)^2}\sqrt{f(s)^2 + g(s)^2}ds - \int_0^t (f(s)f(t)+g(s) g(t))ds\right].$$
By Cauchy-Schwartz inequality, we have
$$\sqrt{f(t)^2 + g(t)^2}\sqrt{f(s)^2 + g(s)^2} \geq f(t)f(s) + g(t) g(s).$$
Hence $\phi'(t) \geq 0$. This implies that $\phi(1) \geq \phi(0) =0$. This is the inequality in question.
In general case when $f$ and $g$ are integrable, we can approximate them by continous functions, and hence finish the proof.