The Question
Suppose that $t$ is a polynomial of degree $n$ or less and satisfies
$$t(x_0)=f_0 \; \; \; \; t(x_n)=f_n \; \; \; \; t(x_i)=0 \; \; , \; \; i=1,2,\dots,n-1$$
where $x_0<x_1<\dots<x_n$.
Let $s$ be the linear spline that interpolates the $t$ at $x_0,x_1,\dots,x_n$. Prove that
$$\int_{x_0}^{x_n}(t'(x))^2dx≥\frac{f_0^2}{x_1-x_0}+\frac{f_n^2}{x_n-x_{n-1}}$$
My Attempt
A hint was given with the question that the identity
$$(t')^2-(s')^2=(t'-s')^2+2s'(t'-s')$$
might be useful. Thus I computed
\begin{align} \ \int_{x_0}^{x_n}(s'(x))^2dx & = \bigl[s(x)s'(x) \bigr]_{x_0}^{x_n}-\int_{x_0}^{x_n}s(x)s''(x)dx \\ \ & = \bigl[s(x_n)s'(x_n)-s(x_0)s'(x_0) \bigr]-(0) \\ \ & = t(x_n)\frac{t(x_n)-t(x_{n-1})}{x_n-x_{n-1}}-t(x_0)\frac{t(x_1)-t(x_{0})}{x_1-x_{0}} \\ \ & = \frac{f_0^2}{x_1-x_0}+\frac{f_n^2}{x_n-x_{n-1}} \end{align}
This seems promising, but unfortunately I have no idea where to go from here. Any hints?
OK, nobody answered my question, but I have finally figured it out anyway, so I shall answer my own question (just talking to myself here).
First we compute
$$\int_{x_0}^{x_n}s'(x)\bigl(t'(x)-s'(x) \bigr)dx = \Bigl[s'(x)\bigl(t(x)-s(x)\bigr) \Bigr]_{x_0}^{x_n}-\int_{x_0}^{x_n}s''(x)\bigl(t(x)-s(x) \bigr)dx = 0 $$
since $t(x_n)=s(x_n)=f_n$ and $t(x_0)=s(x_0)=f_0$; and that $s$ is piecewise linear, so $s''$ is identically $0$. Using this result, we get
\begin{align} \ \int_{x_0}^{x_n}\bigl((t'(x))^2-(s'(x))^2\bigr)dx & = \int_{x_0}^{x_n}\bigl(t'(x)-s'(x) \bigr)^2dx+2\int_{x_0}^{x_n}s'(x)\bigl(t'(x)-s'(x) \bigr)dx \\ \ & = \int_{x_0}^{x_n}\bigl(t'(x)-s'(x) \bigr)^2dx+2(0) \\ \ & = \int_{x_0}^{x_n}\bigl(t'(x)-s'(x) \bigr)^2dx≥0 \end{align}
Hence the result follows:
$$\int_{x_0}^{x_n}(t'(x))^2dx≥\int_{x_0}^{x_n}(s'(x))^2dx=\frac{f_0^2}{x_1-x_0}+\frac{f_n^2}{x_n-x_{n-1}}$$