Answer
(a) $|\langle f,g \rangle|=\sqrt {1-\frac{2}{e}}\leq\| f \|\| g \|=\sqrt{\frac{1}{3}}\sqrt{\frac{1}{2}(1-\frac{1}{e^2})}$
(b) $\| f+g \| =\sqrt{\frac{17}{6}-\frac{1}{2e^{2}}-\frac{4}{e}} \leq\| f \|+\| g \|=\sqrt{\frac{1}{3}}+\sqrt{\frac{1}{2}(1-\frac{1}{e^2})}$.
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
$\langle f, g\rangle=\int_{0}^1 xe^{-x} d x=\left[-xe^{-x}-e^{-x}\right]_{0}^1=1-\frac{2}{e},$
$\langle f, f\rangle=\int_{0}^{1} x^2 d x= \frac{1}{3}\left[x^3\right]_{0}^{1}=\frac{1}{3},$
$\langle g, g\rangle=\int_{0}^{1}e^{-2x}d x=- \frac{1}{2}\left[e^{-2x}\right]_{0}^{1}=\frac{1}{2}(1-\frac{1}{e^2}),$
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{17}{6}-\frac{1}{2e^{2}}-\frac{4}{e}\end{aligned}.
Since for any $f$ we have $\| f \| =\sqrt{\langle f, f\rangle} $, then
(a) $|\langle f,g \rangle|=\sqrt {1-\frac{2}{e}}\leq\| f \|\| g \|=\sqrt{\frac{1}{3}}\sqrt{\frac{1}{2}(1-\frac{1}{e^2})}$
(b) $\| f+g \| =\sqrt{\frac{17}{6}-\frac{1}{2e^{2}}-\frac{4}{e}} \leq\| f \|+\| g \|=\sqrt{\frac{1}{3}}+\sqrt{\frac{1}{2}(1-\frac{1}{e^2})}$.
