real analysis – Show that $int^b_a f^3(x)dx le ( int^b_a f(x)dx )^2$ when $f(a)=0$ and $0 le f'(x) le 1$

Assume that $f:(a,b) to mathbb R$ is a continuously differentiable function. We want to show that $int^b_a f^3(x)dx le ( int^b_a f(x)dx )^2$ if we have $f(a)=0$ and $0 le f'(x) le 1$ for all $x in (a,b)$.

My attempt:

I don’t know if my approach is correct, but I started by changing the upper limits of the integrals to a variable $y$ so that I can differentiate and obtain $f’$. After changing the upper limits, I want to show

$$( int^y_a f(x)dx )^2 – int^y_a f^3(x)dx ge 0$$

for $y in (a,b)$. This holds for $y=a$. So I will now show that the derivative of LHS w.r.t. $y$ is positive. That is, I want to show

$$2f(y)(int^y_a f(x)dx) – f^3(y) ge 0$$

Again, this holds for $y=a$ since $f(a)=0$. And I take the derivative again and want to show that it is positive. The derivative is

$$2f'(y)(int^y_a f(x)dx) +2f^2(y)-3f^2(y)f'(y)$$

I couldn’t get rid of the integral and I am stuck here. Can you help me by proposing another approach or giving a solution?