Consider the vector space of polynomials over $\mathbb{R}$. Then: \begin{equation} \langle p, q \rangle = \int_{-1}^1 p(x)q(x) dx \end{equation} is an inner product.
Most of the properties of an inner product are obviously satisfied, for example it's clear that $\langle p + r, q \rangle = \langle p, q \rangle + \langle r, q \rangle$.
I'm struggling to find a neat proof that $\langle p, p \rangle \geq 0$. In general, $p(x) = \sum_{i=0}^\infty p_ix^i$, and so: \begin{equation} p^2(x) = \sum_{i=0}^\infty \left(\sum_{j=0}^i p_j p_{i-j} \right) x^i.\end{equation}
Now, any odd power of $x$ will not contribute to the integral, because (if $k$ is odd): \begin{equation} \int_{-1}^1 x^k dx = \frac{1}{k+1}(1 - 1) = 0. \end{equation}
On the other hand, if $k$ is even: \begin{equation} \int_{-1}^1 x^k dx = \frac{2}{k+1}, \end{equation} so we just need to make sure that: \begin{equation} \sum_{i=0}^\infty \frac{2}{2i+1}\left(\sum_{j=0}^{2i} p_{j} p_{2i-j}\right) \geq 0. \end{equation}
Here I am a bit stuck as I can't see an obvious reason why this is definitely true. Fixing $p$ to be some small degree polynomial it certainly seems true, for example if $p$ has degree $2$ then this becomes: \begin{align} \sum_{i=0}^1 \frac{2}{2i+1} \left(\sum_{j=0}^{2i} p_{j}p_{2i-j}\right) &= 2p_0^2 + \frac{2}{3}(p_0p_2 + p_1^2 + p_2p_0) + \frac{2}{5}(p_2^2) \\ &= \left(2p_0^2 + \frac{2}{3}p_1^2 + \frac{2}{5}p_2^2\right) + \frac{4}{3}p_0p_2 \\ &\geq 2p_0^2 + \frac{2}{5}p_2^2 + \frac{4}{3}p_0p_2 \\ &\geq 0, \end{align} but I can't see an easy method. In fact, to conclude that the above is 0, I resorted to finding the stationary point of $2p_0^2 + \frac{2}{5}p_2^2 + \frac{4}{3}p_0p_2$ and showing that it was a minimum, and $\geq 0$.
I get a strong impression that I'm making this needlessly complicated. How can I argue, in general, that $\langle p, p \rangle \geq 0$?
You are indeed thinking too hard. Polynomials only have finitely many terms.
Moreover, since $(p(x))^2\ge 0$ for all $x\in[-1,1]$, one has $$ \int_{-1}^1(p(x))^2\;dx\ge 0 $$