$f$ convex, $g$ concave and increasing, $\int_0^1 f = \int_0^1 g$, then $\int_0^1(f)^2 \geq \int_0^1(g)^2$ 
Let $f,g:[0,1] \to [0, \infty)$ be two continuous functions such that $$f(0) = g(0) = 0,$$ $f$ is convex, $g$ is concave and increasing and $$\displaystyle \int_0^1f(x)dx = \int_0^1g(x)dx.$$ Prove that $$ \displaystyle \int_0^1\left(f(x) \right)^2dx \geq \int_0^1\left(g(x) \right)^2dx.$$

I don't quite know how to approach the problem. 
I thought about using Chebyshev's inequality due to the fact that $g$ is increasing, $F(x) = \int_0^xf(t)dt$ is increasing (since $f$ is convex) and $G(x) = \int_0^xg(t)dt$ is increasing (since $g$ is increasing and $g(0) = 0$), but it didn't help me. 
I also tried to obtain something by writing the convexity and concavity point-wise and using the fact that $\displaystyle h(x) = \frac{f(x)}{x}$ is increasing and $p(x) = \displaystyle \frac{g(x)}{x}$ is decreasing, but I got nothing.

For some $0 < \epsilon < 1$,
$$
\int_\epsilon^1 (f(x)^2-g(x)^2)dx = \int_\epsilon^1 (f(x) + g(x))(f(x)-g(x)) dx
$$
since $f(x)+g(x) \ge 0$, we can apply the intermediate value theorem to get a $\xi \in [\epsilon,1]$, such that
$$
\int_\epsilon^1 (f(x) + g(x))(f(x)-g(x))dx = (f(\xi)+g(\xi))\int_\epsilon^1 (f(x)-g(x))dx \,.
$$
Since $f(\xi)+g(\xi) \ge 0$, it remains to show that 
$$
\int_\epsilon^1 (f(x)-g(x))dx \ge 0
$$
$h(x) = (f(x)-g(x))$ is also convex ($-g(x)$ is convex) with ($h(0) = 0$), so $h$ can only have at most 2 roots (except $h \equiv 0$). If there is no other root $x_0 \in(0,1)$, then either $f(x) > g(x)$ or the other way around, and the integrals would not match. So there must be $x_0 \in(0,1)$ such that $h(x) \ge 0$ for all $x \ge x_0$. Therefore $f(x) \ge g(x)$ and furthermore $\int_\epsilon^1(f(x)-g(x))dx \ge 0$, with $\epsilon := x_0$.
