Answer
(a) $|\langle u, v \rangle|=2\leq\| u \|\| v \|=(\sqrt 2)(\sqrt 2)=2 $
(b) $\| u+v \| =0\leq \| u \|+\| v \|=2\sqrt 2$.
Work Step by Step
Let $u=(-1,1), \quad v=(1,-1)$, $\langle u, v \rangle=u\cdot v$, then we have
$\langle u, v \rangle = u_1v_1+u_2v_2=-1-1=-2$,
$\| u \| =\sqrt{\langle u, u\rangle}=\sqrt{u_1^2+u_2^2}=\sqrt{2}$,
$\| v \| =\sqrt{\langle v, v\rangle}=\sqrt{v_1^2+v_2^2}=\sqrt{2}$,
$\| u+v \| =\| (0,0) \| =0$,
Now, we get
(a) $|\langle u, v \rangle|=2\leq\| u \|\| v \|=(\sqrt 2)(\sqrt 2)=2 $
(b) $\| u+v \| =0\leq \| u \|+\| v \|=2\sqrt 2$.
