Answer
See the explanation below.
Work Step by Step
Let $f(x)=\sin x, \quad g(x)=\cos x.$, $\langle f, g\rangle=\int_{0}^{\frac{\pi}{4}}f(x)g(x) d x$, we have $\langle f, g\rangle=\int_{0}^{\frac{\pi}{4}}\sin x\cos x d x=\frac{1}{2}\left[\sin^{2}x\right]_{0}^{\frac{\pi}{4}}=\frac{1}{4},$ $\langle f, f\rangle=\int_{0}^{\frac{\pi}{4}}\sin^2 x d x=\frac{1}{2}\int_{0}^{\frac{\pi}{4}}(1-\cos 2x) d x=\frac{1}{2}\left[x-\frac{1}{2}\sin 2x\right]_{0}^{\frac{\pi}{4}}=\frac{1}{2}(\frac{\pi}{4}-\frac{1}{2}),$ $\langle g, g\rangle=\int_{0}^{\frac{\pi}{4}}\cos^2 x d x=\frac{1}{2}\int_{0}^{\frac{\pi}{4}}(1+\cos 2x) d x=\frac{1}{2}\left[x+\frac{1}{2}\sin 2x\right]_{0}^{\frac{\pi}{4}}=\frac{1}{2}(\frac{\pi}{4}+\frac{1}{2}),$ \begin{aligned}\langle f+g, f+g\rangle &=\int_{0}^{\frac{\pi}{4}}(\sin x +\cos x)^2d x\\ &=\int_{0}^{\frac{\pi}{4}}\left(1+\sin 2x\right) d x \\ &=\left[x-\frac{1}{2}\cos 2x\right]_{0}^{\frac{\pi}{4}} \\ &=\frac{\pi}{4}+\frac{1}{2}\end{aligned}
Since for any $f$, we have $\| f \| =\sqrt{\langle f, f\rangle} $, then:
(a) Cauchy-Schwarz Inequality: $|\langle f,g \rangle|=\frac{1}{4}\leq\| f \|\| g \|=\frac{1}{2}\sqrt{\frac{\pi^2}{16}-\frac{1}{4}}$
(b) The triangle inequality: $\| f+g \| =\sqrt{\frac{\pi}{4}+\frac{1}{2}}\leq\| f \|+\| g \|=\frac{1}{\sqrt 2}{\sqrt{\frac{\pi}{4}-\frac{1}{2}}}+\frac{1}{\sqrt 2}{\sqrt{\frac{\pi}{4}+\frac{1}{2}}}$
