Show that $\left(\int_{0}^{1}\sqrt{f(x)^2+g(x)^2}\ dx\right)^2 \geq \left(\int_{0}^{1} f(x)\ dx\right)^2 + \left(\int_{0}^{1} g(x)\ dx\right)^2$ 
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.
 A: Let a curve $C\in\mathbb{R}^2$ be defined by the parametrisation $\displaystyle x(t)=x(0)+\int_0^t f(x)dx$ and $\displaystyle y(t)=y(0)+\int_0^t g(x)dx$, $t\in [0, 1]$. Then the LHS is the square of the arc length of $C$ joining $(x(0), y(0))$ and $(x(1), y(1))$, whereas the RHS is the square of the shortest distance between $(x(0), y(0))$ and $(x(1), y(1))$. 
A: Definition: $u:\mathbb{R}^n\to\mathbb{R}$ is convex if for all $a\in\mathbb{R}^n$, there is a $v(a)\in\mathbb{R}^n$ so that for all $x\in\mathbb{R}^n$
$$
u(x)-u(a)\ge v(a)\cdot(x-a)\tag{1}
$$
Theorem (Extension of Jensen): If $u:\mathbb{R}^n\to\mathbb{R}$ is convex, $f:\Omega\to\mathbb{R}^n$, and $\int_\Omega\mathrm{d}\omega=1$, then 
$$
\int_\Omega u(f)\,\mathrm{d}\omega\ge u\left(\int_\Omega f\,\mathrm{d}\omega\right)\tag{2}
$$
Proof: Let $a=\int_\Omega f\,\mathrm{d}\omega$. Then $(1)$ becomes
$$
u(f)-u\left(\int_\Omega f\,\mathrm{d}\omega\right)
\ge v\left(\int_\Omega f\,\mathrm{d}\omega\right)
\cdot\left(f-\int_\Omega f\,\mathrm{d}\omega\right)\tag{3}\\
$$
Since $\int_\Omega\mathrm{d}\omega=1$, integrating $(3)$ over $\Omega$ gives
$$
\begin{align}
\int_\Omega u(f)\,\mathrm{d}\omega-u\left(\int_\Omega f\,\mathrm{d}\omega\right)
&\ge v\left(\int_\Omega f\,\mathrm{d}\omega\right)
\cdot\left(\int_\Omega f\,\mathrm{d}\omega-\int_\Omega f\,\mathrm{d}\omega\right)\\[3pt]
&=0\tag{4}
\end{align}
$$
QED
Claim: $u(x)=\left\|x\right\|$ is convex.
Proof:
$$
\begin{align}
\left\|a\right\|\left\|x\right\|&\ge a\cdot x\tag{5}\\[6pt]
\left\|x\right\|&\ge\frac{a}{\left\|a\right\|}\cdot x\tag{6}\\
\left\|x\right\|-\left\|a\right\|&\ge\frac{a}{\left\|a\right\|}\cdot(x-a)\tag{7}
\end{align}
$$
Explanation:
$(5)$: Cauchy-Schwarz
$(6)$: divide both sides by $\left\|a\right\|$
$(7)$: subtract $\left\|a\right\|$ from both sides
QED
The theorem and the claim prove that
$$
\int_0^1\left\|f(x)\right\|\mathrm{d}x\ge\left\|\int_0^1f(x)\,\mathrm{d}x\right\|\tag{8}
$$
which, in $\mathbb{R}^2$, is the inequality in the question.
A: 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.
