Answer
(a) $|\langle f,g \rangle|=1\leq\| f \|\| g \|=\sqrt{\frac{1}{3}}\sqrt{\frac{1}{2}(e^2-1)}$
(b) $\| f+g \| =\sqrt{\frac{11}{6}+\frac{e^2}{2}} \leq\| f \|+\| g \|=\sqrt{\frac{1}{3}}+\sqrt{\frac{1}{2}(e^2-1)}$.
Work Step by Step
Let $f(x)=x, \quad g(x)=e^x$, $\langle f, g\rangle=\int_{0}^1 f(x)g(x) d x$, we have
$\langle f, g\rangle=\int_{0}^1 xe^x d x=\left[xe^x-e^x\right]_{0}^1=1,$
$\langle f, f\rangle=\int_{0}^{1} x^2 d x= \frac{1}{3}\left[x^3\right]_{0}^{1}=\frac{1}{3},$
$\langle g, g\rangle=\int_{0}^{1}e^{2x}d x= \frac{1}{2}\left[e^{2x}\right]_{0}^{1}=\frac{1}{2}(e^2-1),$
by making use of the calculations above, we get
\begin{aligned}\langle f+g, f+g\rangle &=\int_{0}^{1}( x +e^{x})^2d x\\ &=\int_{0}^{1}\left(x^2+e^{2x}+2x e^{x} \right) d x \\
&=\left[\frac{1}{3}x^3+\frac{1}{2}e^{2x}+2 (xe^x-e^x)\right]_{0}^{1} \\ &=\frac{11}{6}+\frac{e^2}{2}\end{aligned}.
Since for any $f$ we have $\| f \| =\sqrt{\langle f, f\rangle} $, then
(a) $|\langle f,g \rangle|=1\leq\| f \|\| g \|=\sqrt{\frac{1}{3}}\sqrt{\frac{1}{2}(e^2-1)}$
(b) $\| f+g \| =\sqrt{\frac{11}{6}+\frac{e^2}{2}} \leq\| f \|+\| g \|=\sqrt{\frac{1}{3}}+\sqrt{\frac{1}{2}(e^2-1)}$.