Answer
(a) $\langle f, g\rangle= 0,$
(b) Since $\langle f, g\rangle=0$, then $f$ and $g$ are orthogonal.
(c) see details below.
Work Step by Step
Let $f(x)=x, \quad g(x)=\frac{1}{x^2+1}.$, $\langle f, g\rangle=\int_{-1}^1 f(x)g(x) d x$, we have
(a) $\langle f, g\rangle=\int_{-1}^1 \frac{x}{x^2+1} d x=\frac{1}{2}\left[\ln \frac{1}{x^2+1}\right]_{-1}^1=0,$
(b) Since $\langle f, g\rangle=0$, then $f$ and $g$ are orthogonal.
(c) To verify the Cauchy-Schwarz Inequality, we have
$$\langle f,f\rangle=\int_{-1}^1x^2 d x=\frac{2}{3},$$
by using the substitution $x=\tan u$ one can have the follwoing
$$\langle g, g\rangle=\int_{-1}^1 \frac{1}{(x^2+1)^2} d x=\frac{\pi+2}{4} .$$
Since for any $f$ we have $\| f \| =\sqrt{\langle f, f\rangle} $, then we have the Cauchy-Schwarz Inequality
$$|\langle f,g \rangle|=0\leq\| f \|\| g \|=\sqrt{\frac{2}{3}}\sqrt{\frac{\pi+2}{4} }$$
