Proving that a given mapping is a inner product How to prove that a mapping $\langle\cdot,\cdot\rangle:\mathbb{R_{3}[x]}\times\mathbb{R_{3}[x]}\rightarrow\mathbb{R}$ is a inner product, where $\langle f,g\rangle=\int\limits_{-1}^{1}f(x)g(x)dx?$
First, for $a,b\in\mathbb{R}$ and $f,g,h\in\mathbb{R_{3}[x]}$, we have
$\langle af+bg,h\rangle=\int\limits_{-1}^{1}(af+bg)(x)h(x)dx=\int\limits_{-1}^{1}(af(x)+bg(x))h(x)dx$
$\qquad\qquad\quad\  =a\int\limits_{-1}^{1}f(x)h(x)dx+b\int\limits_{-1}^{1}g(x)h(x)dx=a\langle f,h\rangle+b\langle g,h\rangle.$
Second, for $f,g\in\mathbb{R_{3}[x]}$, we have
$\langle f,g\rangle=\int\limits_{-1}^{1}f(x)g(x)dx=\int\limits_{-1}^{1}g(x)f(x)dx=\langle g,f\rangle$.
And, third, for $f\in\mathbb{R}_3[x],$ $f\neq0,$ we have to prove $\langle f,f\rangle >0.$
I wrote $f=ax^3+bx^2+cx+d$ and I get $\langle f,f\rangle=\dfrac{2}{7}a^2+\dfrac{4}{5}ac+\dfrac{2}{3}c^2+\dfrac{2}{5}b^2+\dfrac{4}{3}bd+2d^2.$
I don't know how to prove that this is greater than 0. I tried to make a square of a binomial, but it isn't working.

The answers given already are the most elegant, but for completeness let me give another way (continuing the approach in the post). Let $f = ax^3 +b x^2+cx +d.$ Then
\begin{align*}
\langle f, f \rangle &= \frac{2}{7} a^2 + \frac{4}{5} ac + \frac{2}{3} c^2 + \frac{2}{5} b^2 + \frac{4}{3} bd + 2d^2 \\
&= \frac{2}{3} \left( c + \frac{3a}{5}\right)^2 + \frac{8}{175} a^2 + 2 \left(d+\frac{b}{3}\right)^2 + \frac{8}{45} b^2,
\end{align*}
which is greater than zero (and equal to zero iff $f=0$).
