Answer
See the explanation below.
Work Step by Step
Let $f(x)=x, \quad g(x)=\cos \pi x.$, $\langle f, g\rangle=\int_{0}^2 f(x)g(x) d x$, we have $\langle f, g\rangle=\int_{0}^2 x\cos \pi x d x=\left[\frac{ x}{ \pi }\sin \pi x+\frac{ 1}{ \pi^2 }\cos \pi x\right]_{0}^2=0,$ $\langle f, f\rangle=\int_{0}^{2} x^2 d x= \frac{1}{3}\left[x^3\right]_{0}^{2}=\frac{8}{3},$ $\langle g, g\rangle=\int_{0}^{2}\cos^2 \pi x d x=\frac{1}{2}\int_{0}^{2}(1+\cos 2\pi x) d x=\frac{1}{2}\left[x+\frac{1}{2\pi}\sin 2\pi x\right]_{0}^{2}=1,$
By making use of the calculations above, we get
\begin{aligned}\langle f+g, f+g\rangle &=\int_{0}^{2}( x +\cos \pi x)^2d x\\ &=\int_{0}^{2}\left(x^2+\cos^2 \pi x+2x \cos \pi \right) d x \\ &=\left[\frac{1}{3}x^3+\frac{1}{2}(x+\frac{1}{2\pi}\sin 2\pi x)\right]_{0}^{2} \\ &=\frac{11}{3}\end{aligned}
Since for any $f$, we have $\| f \| =\sqrt{\langle f, f\rangle} $, then
(a) Cauchy-Schwarz Inequality: $|\langle f,g \rangle|=0\leq\| f \|\| g \|=\sqrt{\frac{8}{3}}$
(b) The triangle inequality: $\| f+g \| =\sqrt{\frac{11}{3}} =1.91\leq\| f \|+\| g \|=\sqrt{\frac{8}{3}}+1=2.63$.
