How to prove $\int_0^1 f^2(x)\cdot f'^4(x)\ dx\leq \int_0^1 f^4(x)\cdot f''^2(x)\ dx$ [closed]

Suppose $f(x) \in C^2[0,1]$ with $f(x)>0\quad\forall x\in [0,1]$. If $f'(0)=f'(1)=0$,¿ how to prove $$\int_0^1 f^2(x)\cdot (f')^4(x)\ dx\leq \int_0^1 f^4(x)\cdot (f'')^2(x)\ dx$$

Using the integration by parts hint given in the comments and the fact that $f'(0)=f'(1)=0$ we obtain

$$\int_0^1 f^2(x)(f'(x))^4 dx = - \int_0^1 f^3(x) (f'(x))^2 f''(x) dx\ .$$

Note that the left hand side is nonnegative. Then apply Cauchy Schwarz

$$\begin{align}\int_0^1 f^2(x)(f'(x))^4 dx & = \left\vert \int_0^1 f^3(x) (f'(x))^2 f''(x) dx\right\vert \\ &= \left\vert \int_0^1 \left[f^2(x)f''(x)\right]\left[f(x) (f'(x))^2\right] dx\right\vert \\ &\leq \left( \int_0^1 \left\vert f^2(x)f''(x)\right\vert^2 dx\right)^{1/2}\left(\int_0^1\left\vert f(x) (f'(x))^2\right\vert^2 dx\right)^{1/2}\\ &\leq \left( \int_0^1 f^4(x)(f''(x))^2 dx\right)^{1/2}\left(\int_0^1 f^2(x) (f'(x))^4 dx\right)^{1/2}\ . \end{align}$$

Dividing by the term on the right and squaring gives

$$\int_0^1 f^2(x)(f'(x))^4 dx \leq \int_0^1 f^4(x)(f''(x))^2 dx$$