Let $\displaystyle\;\;\lambda = \frac{\pi}{b-a}.\;\;$ Since $f(a) = f(b) = 0$ and $f \in C^1[a,b]$, the function defined by:
$$\varphi(x) = \begin{cases}
\dfrac{f'(a)}{\lambda}, & x = a\\
\\
\\
\dfrac{f(x)}{\sin(\lambda(x-a))}, & a < x < b\\
\\
\\
-\dfrac{f'(b)}{\lambda}, & x = b
\end{cases}$$
is $C^1$ over $(a,b)$ and continuous at $a$ and $b$. We have
$$\int_a^b |f'(x)|^2 dt
= \int_a^b \Big(\varphi'(x) \sin(\lambda(x-a)) + \lambda \varphi(x) \cos(\lambda(x-a))\Big)^2 dx
$$
Notice the cross term in the integrand can be simplified as:
$$\begin{align}
& 2\lambda \varphi(x)\varphi'(x) \sin(\lambda(x-a))\cos(\lambda(x-a))\\
= & \frac{\lambda}{2} (\varphi^2(x))'\sin(2\lambda(x-a))\\
= & \frac{\lambda}{2} \frac{d}{dx}\left[ \varphi^2(x) \sin(2\lambda(x-a))\right] - \lambda^2 \varphi^2(x) \cos(2\lambda(x-a))
\end{align}$$
We find
$$\begin{align}
& \int_a^b |f'(x)|^2 dt\\
= & \int_a^b \Big( |\varphi'(x)|^2 + \lambda^2 |\varphi(x)|^2 \Big) \sin^2(\lambda(x-a)) dx
+ \frac{\lambda}{2} \left[\varphi^2(x) \sin(2\lambda(x-a))\right]_a^b\\
\ge & \lambda^2 \int_a^b |\varphi(x)|^2 \sin^2(\lambda(x-a)) dx\\
= & \lambda^2 \int_a^b |f(x)|^2 dx\\
\ge & \frac{1}{(b-a)^2} \int_a^b |f(x)|^2 dx
\end{align}
$$
Please note that above steps actually contain a proof of Wirtinger's inequality for functions mentioned in Noam's answer. For other proofs of this inequality, please see this
question and the links there.
