Prove that $\int_0^\infty xf(x) dx = a$ and $0<f<1$ implies $\int_0^\infty xf(x) dx > \frac{a^2}{2}$

Question

Suppose that $f \in C[0,\infty), 0 < f(x) < 1, \displaystyle \int_0^\infty xf(x) dx = a$. Proof that $$\displaystyle \int_0^\infty xf(x) dx > \dfrac{a^2}{2}$$ Here is my attempt: I try to proof that $$\displaystyle \int_0^\infty xf(x) dx > \displaystyle \int_0^a x dx $$ but I got stuck