Elementary Linear Algebra 7th Edition

(a) $|\langle u, v \rangle|=2\leq\| u \|\| v \|=(\sqrt 2)(\sqrt 2)=2$ (b) $\| u+v \| =0\leq \| u \|+\| v \|=2\sqrt 2$.
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) $\| u+v \| =0\leq \| u \|+\| v \|=2\sqrt 2$.