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$\forall x(t)\in C^1[a,b],x(a)=0,\int^b_a \dot x^2(t) dt\geq \frac{\pi ^2}{4(b-a)^2}\int^b_a x^2(t)dt$

I tried to prove this by following steps:

(1)let $k^2=\frac{\pi ^2}{4(b-a)^2}$,for $x(t)\in C^2[a,b],x(b)=B$, solve E-L equations and we get $ \ddot x(t)+k^2x=0$, so $x(t)=sin(k(x-a)+\phi)-sin\phi$, so that the proposition right.

(2) for $x(t)\in C^1[a,b]$,we can use function series in$C^2$ to approximate $x(t)$.

The problem is, E-L equations only gives the local maximum value,so this proof actually doesn't work.

So I wanna know if the method above can be improve, or if there are other elementary methods(maybe without using Fourier series or too advanced functional analysis, only the undergraduate Mathematical analysis level) to solve this problem.

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WLOG: $a=0, b=\pi/2$.
We have: $$\begin{align} \int_0^{\pi/2} x^2(t)dt &= \int_0^{\pi/2} \left( \int_0^{t} \dot{x}(s)ds \right)^2dt \\ &\stackrel{ \text{Holder}}{\le}\int_0^{\pi/2} \left( \int_0^{t} \dot{x}^2(s)\frac{1}{\cos (s)}ds \right)\left( \int_0^t \cos(s)ds \right) dt \\ &= \int_0^{\pi/2} \int_0^t \dot{x}^2(s)\frac{\sin(t)}{\cos (s)}dsdt \\ &\stackrel{\text{Fubini}}{=} \int_0^{\pi/2}\int_{s}^{\pi/2}\dot{x}^2(s)\frac{\sin(t)}{\cos (s)}dtds \\ &= \int_0^{\pi/2}\dot{x}^2(s)ds \end{align}$$ QED.

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