Alternative proof of simple integral inequality

Problem. Let $f\in C^1(\mathbb R)$ such that $f(0) = 0$ and $0 < f'(x) \le 1$. Prove that for all $x\ge 0$ $$ \int_0^x f^3(t)\,dt \le \left(\int_0^x f(t)\,dt\right)^{\!\!2}. $$

Below is my solution of this simple problem, but I want to see alternative proof (just for fun). Thank you for your time!

Since $f'(x) > 0$ then $f(x) > 0$ while $x>0$. $f'(x)\le 1$ is given; let's multiply it by $2f(x)$: $$ 2f(x)f'(x) \le 2f(x). $$ Let's integrate this inequality from $0$ to $x$: $$ \int_0^x 2f(t)f'(t)\,dt \le \int_0^x 2f(t)\,dt, $$ but $$ \int_0^x 2f(t)f'(t)\,dt = \int_0^x \frac{d}{dt} \Big(f^2(t)\Big)dt = f^2(x) - f^2(0) = f^2(x). $$ So, $$ f^2(x) \le 2\int_0^x f(t)\,dt $$ Let's multiply it by $f(x) > 0$: $$ f^3(x) \le 2f(x)\int_0^x f(t)\,dt, $$ and integrate: $$ \int_0^x f^3(t)\,dt \le \int_0^x 2f(t)\left(\int_0^t f(u)\,du\right)dt $$ Denote $F(t) = \displaystyle\int_0^t f(u)\,du$. Then RHS is $$ \int_0^x 2f(t)F(t)\,dt = \int_0^x 2F'(t)F(t)\,dt = \int_0^x \frac{d}{dt} \Big(F^2(t)\Big)dt = F^2(x) - F^2(0) = F^2(x), $$ or $$ \int_0^x f^3(t)\,dt \le \left(\int_0^x f(t)\,dt\right)^2. $$

EDIT

Solution of @J. J. is pretty and correct, but it's not so alternative. In other words, we prove that $f(x)-g(x)\ge 0$ instead of $f(x)\ge g(x)$. I want to see proof with known integral inequalities (like Cauchy-Schwarz, for example).

Let $$h(x) = \left(\int_0^x f(t) \, dt\right)^2 - \int_0^x f^3(t) \, dt.$$ We need to show that $h(x) \ge 0$ for $x \ge 0$. Clearly $h(0) = 0$ so it is enough to show that $h'(x) \ge 0$ for all $x \ge 0$. We have $$h'(x) = 2 \int_0^x f(t) \, dt f(x) - f^3(x) = f(x) \left( 2 \int_0^x f(t) \, dt - f^2(x) \right),$$ so it is enough to show that $g(x) := 2 \int_0^x f(t) \, dt - f^2(x) \ge 0$ for all $x \ge 0$. Now clearly $g(0) = 0$ so it is enough to show that $g'(x) \ge 0$ for all $x \ge 0$. We have $$g'(x) = 2 f(x) - 2 f(x) f'(x) = 2f(x)(1 - f'(x)) \ge 0,$$ which proves the claim.