Answer
(a) $|\langle p, q \rangle|=0\leq\| p \|\| q \|=2$
(b) $\| p+q \| =2\leq\| p \|+\| q \|=2\sqrt {2}.$
Work Step by Step
Let $p(x)=x, \quad q(x)=1-x^2, \quad \langle p, q \rangle = a_0b_0+2a_1b_1+a_2b_2$, we have
$\langle p, q \rangle = a_0b_0+2a_1b_1+a_2b_2=0+0+0=0$,
$\| p \| =\sqrt{\langle p, p\rangle}=\sqrt{a_0^2+2a_1^2+a_3^2}=\sqrt{2}$,
$\| q \| =\sqrt{\langle q, q\rangle}=\sqrt{b_0^2+2b_1^2+b_3^2}=\sqrt{1+0+1}=\sqrt{2}$,
$\| p+q \|=\| 1+x-x^2 \|=\sqrt{1+2+1}=2$
Now, we get
(a) $|\langle p, q \rangle|=0\leq\| p \|\| q \|=2$
(b) $\| p+q \| =2\leq\| p \|+\| q \|=2\sqrt {2}.$
