Show that $\left[\int_{0}^{1}f(x) \ dx\right]^2\geq\int_{0}^{1}(f(x))^3 \ dx$ Let $f$ be continous function with a continious derivative in $[0,1]$ such that $0<f'(x)<1$ and $f(0)=0.$ Show that $$\left[\int_{0}^{1}f(x) \ dx\right]^2\geq\int_{0}^{1}(f(x))^3 \ dx.$$
I apologize for not having an attempt. Many times I learn the best by observing whole solutions and asking question about them.

We first note that $f(0)=0$ and $f'\geq 0$ imply that for $t\in [0,1]$
$$f(t)=\int_0^t f'(t)dt\geq 0.$$
For $t\in [0,1]$ consider the function
$$H(t):=\left[\int_{0}^{t}f(x) \ dx\right]^2-\int_{0}^{t}(f(x))^3 \ dx$$
Since $H(0)=0$, it suffices to show that $H$ is increasing. Then $H(1)\geq H(0)=0$ and the required inequality holds.
By taking the derivative, we have that
$$H'(t)=2f(t)\int_{0}^{t}f(x) \ dx-f^3(t)=f(t)\left(2\int_{0}^{t}f(x) \ dx-f^2(t)\right).$$
Hence $H$ is increasing iff for $t\in [0,1]$,
$$G(t):=2\int_{0}^{t}f(x) \ dx-f^2(t)\geq 0.$$
Since $f(0)=0$, we have that $G(0)=0$. Moreover $f'\leq 1$ and $f\geq 0$ implies that
$$G'(t):=2f(t) \ dx-2f(t)f'(t)=2f(t)(1-f'(t))\geq 0.$$
Hence $G$ is increasing and $G(t)\geq G(0)=0$.
