Answer
(a) $|\langle p, q \rangle|=0\leq\| p \|\| q \|=2\sqrt {10}$
(b) $\| p+q \| =\sqrt{14}=3.74\leq\| p \|+\| q \|=2+\sqrt {10}=5.16.$
Work Step by Step
Let $p(x)=2x, \quad q(x)=3x^2+1, \quad \langle p, q \rangle = a_0b_0+a_1b_1+a_2b_2$, we have
$\langle p, q \rangle = a_0b_0+a_1b_1+a_2b_2=0+0+0=0$,
$\| p \| =\sqrt{\langle p, p\rangle}=\sqrt{a_0^2+a_1^2+a_3^2}=\sqrt{4}=2$,
$\| q \| =\sqrt{\langle q, q\rangle}=\sqrt{b_0^2+b_1^2+b_3^2}=\sqrt{1+0+9}=\sqrt{10}$,
$\| p+q \|=\| 1+2x+3x^2 \|=\sqrt{1+4+9}=\sqrt{14}$
Now, we get
(a) $|\langle p, q \rangle|=0\leq\| p \|\| q \|=2\sqrt {10}$
(b) $\| p+q \| =\sqrt{14}=3.74\leq\| p \|+\| q \|=2+\sqrt {10}=5.16.$
