= Solution
The form is symmetric and bilinear by the corresponding properties of finite real sums. Its squared <norm> is $\langle p,p\rangle=\sum_i p(x_i)^2\geq0$. Equality means $p$ vanishes at all $n+1$ distinct nodes. A nonzero <polynomial> of degree at most $n$ has at most $n$ roots, so equality forces $p=0$. Thus this is an <inner product> on $\Pi_n$.
On $\Pi_{n+1}$ it is only <positive semidefinite>: the nonzero node <polynomial> $\omega(x)=\prod_{i=0}^n(x-x_i)$ satisfies $\langle\omega,\omega\rangle=0$. Hence \b[it is not an <inner product> on $\Pi_{n+1}$]. The <finite-evaluation inner product on polynomials> works exactly when the evaluation map is injective on the chosen space.
Finally, if $q\ne0$, positivity applied to $p-tq$ and the choice $t=\langle p,q\rangle/\langle q,q\rangle$ give
$$
0\leq\langle p,p\rangle-\frac{\langle p,q\rangle^2}{\langle q,q\rangle}.
$$
Rearranging proves <Cauchy-Schwarz inequality> in its required form:
$$
\boxed{\left(\sum_i p(x_i)q(x_i)\right)^2\leq\left(\sum_i p(x_i)^2\right)\left(\sum_i q(x_i)^2\right)}.
$$
If $q=0$ both sides of the asserted inequality are zero. Equality holds exactly when $p,q$ are <linearly dependent>, since in the nonzero case the minimizing $p-tq$ must have zero <norm>.
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