Suppose $f(x)$ is fourth differentiable and satisfies $f(0)=f(1)=f'(0)=f'(1)=0$.Prove the inequality$$\frac{1}{a^4}\int^{1}_{0}[f''(x)]^2\text dx\geq \int^{1}_{0}[f(x)]^2\text dx$$Where $a$ is the minimum positive root of $\cos a\cosh a=1$;
I tried to use skills of variational method to solve it.I obtained an extreme value of the functional $$L=\frac{1}{a^2}[f''(x)]^2-[f(x)]^2$$
by applying Euler-Lagrange equation.$$f(x)=C_1\cos ax+C_2 e^{-ax}+C_3\sin ax+C_4 e^{ax}$$ is its extreme value,where the constants satisfies a system of linear equations$$\begin{cases}C_2-C_3-C_4=0\\C_1+C_2+C_4=0\\C_1\sin a+C_2 e^{-a}-C_3\cos a-C_4 e^a=0\\ C_1\cos a+C_2 e^{-a}+C_3\sin a+C_4 e^a=0\end{cases}$$ because the boundary conditions of $f$.
But I still can't prove the inequality because I don't know the sufficient condition of minimizing the functional$$\int^{b}_{a}L(x,y,y'')\text dx$$(I only find something about the sufficiency of minimizing the functional $\displaystyle{\int^{b}_{a}L(x,y,y')\text dx}$)
I also tried to aviod this problem by setting $f(x)=h(x)g(x)$ and I obtain $$\begin{align}L&=\int^{1}_{0}h(x)^2g''(x)^2+g(x)^2h''(x)^2+4h'(x)^2g'(x)^2-a^4h(x)^2g(x)^2\\&\quad+4g(x)h'(x)g'(x)h''(x)+4h'(x)h(x)g''(x)g'(x)\\&\quad+2h(x)g(x)h''(x)g''(x)\text dx&\end{align}$$where $h(x)$ is the extreme value of the functional.Using Integration by parts,the equation can be simplified to$$\begin{align}L&=\int^{1}_{0}h(x)^2g''(x)^2\text dx+2\int^{1}_{0}h'(x)^2g'(x)^2\text dx\\&\quad-4\int^{1}_{0}h(x)h''(x)g'(x)^2\text dx\end{align}$$But it seems that $h'(x)^2-2h(x)h''(x)\geq0$ is wrong when $x$ is very close to zero.
If anyone gives me some advice on it,I'll be very grateful.