Let $|\langle u,v\rangle|=\|u\| \cdot \|v\|.$ How to show that $u,v$ are linearly dependent? 
Let $|\langle u,v\rangle|=\|u\|\cdot \|v\|.$ How to show that $u,v$ are linearly dependent?

Without loss of generality let $u\ne0,v\ne0.$ 
Then, $\displaystyle\left\langle\frac{u}{\|u\|},\frac{v}{\|v\|}\right\rangle=1=\left\langle\frac{u}{\|u\|},\frac{u}{\|u\|}\right\rangle\\\implies\displaystyle\left\langle\frac{u}{\|u\|},\frac{u}{\|u\|}-\frac{v}{\|v\|}\right\rangle=0$
I don't know what to do next.

Added: A thought that just occurred to me:
$\displaystyle\left\langle\frac{u}{\|u\|},\frac{u}{\|u\|}-\frac{v}{\|v\|}\right\rangle=0=\left\langle \frac{v}{\|v\|},0\right\rangle\\\implies \displaystyle\left\langle\frac{u}{\|u\|}-\frac{v}{\|v\|},\frac{u}{\|u\|}-\frac{v}{\|v\|}\right\rangle=0\\\implies \dfrac{u}{\|u\|}-\dfrac{v}{\|v\|}=0$
Does it work?
 A: Write $\lambda=\langle u,v\rangle/\Vert v\Vert^2$.  Let 
$$
w=u-\lambda v.
$$
Then
$$
\begin{aligned}
\langle w,w\rangle&=\langle u,u\rangle-\overline{\lambda}\langle u,v\rangle-\lambda\langle v, u\rangle+|\lambda|^2\langle v,v\rangle\\
&=\Vert u\Vert^2-2\frac{|\langle u,v\rangle|^2}{\Vert v\Vert^2}+\frac{|\langle u,v\rangle|^2}{\Vert v\Vert^4}\Vert v\Vert^2\\
&=\Vert u\Vert^2-2\frac{\Vert u\Vert^2\cdot\Vert v\Vert^2}{\Vert v\Vert^2}+
\frac{\Vert u\Vert^2\cdot\Vert v\Vert^2}{\Vert v\Vert^2}\\
&=\Vert u\Vert^2(1-2+1)=0.
\end{aligned}
$$
So $w=0$ and $u$ and $v$ must be dependent.
A: If we are talking about a real inner product,
$$
\langle\,u,v\,\rangle=\pm\|u\|\|v\|
$$
implies
$$
\langle\,u\|v\|\mp v\|u\|,u\|v\|\mp v\|u\|\,\rangle=0
$$
Therefore, $u\|v\|\mp v\|u\|=0$.

If we are talking about a complex inner product, then
$$
|\langle\,u,v\,\rangle|=\|u\|\|v\|
$$
We can find a $z\in\mathbb{C}$ so that $|z|=1$ and $\langle\,zu,v\,\rangle\ge0$ (that is, it is a non-negative real).
$$
\begin{align}
\langle\,zu\|v\|-v\|u\|,zu\|v\|-v\|u\|\,\rangle
&=2\|u\|^2\|v\|^2-2\mathrm{Re(}\langle\,zu,v\,\rangle)\|u\|\|v\|\\[6pt]
&=0
\end{align}
$$
A: An idea: let $\,x\in\Bbb R\,$ be a parameter, then
$$ \langle u+xv,u+xv\rangle=||u||^2+2\text{Re}\langle u,v\rangle x+x^2||v||^2\;\;\;\;(**)$$
The above real quadratic in $\,x\,$ has a unique root iff
$$\Delta:=b^2-4ac=4\left(\text{Re}\langle u,v\rangle\right)^2-4||u||^2\,||v||^2=0\iff\left|\text{Re}\langle u,v\rangle\right|=||u||\,||v||$$
In this case the quadratic is 
$$||v||^2x^2\pm 2||u||\,||v||\,x+||u||^2 =\left(||v||x\pm||u||\right)^2$$
and the unique (double root) is
$$x=\pm \frac{||u||}{||v||}$$
and thus (**) becomes
$$0=\langle u+xv,u+xv\rangle\iff u=-xv\iff\;u\;,\;v\;\;\text{are linearly dependent}$$
A: Suppose $u=kv$ for some $k\neq0$ what will you get?
A: Suppose $u\neq 0$ and $v\neq 0$ are linearly dependent, i.e. exist coefficients $ \lambda, \gamma$ not both equal to $0$ such that $w:=\lambda u+\gamma v=0$. 
But then
$$0=|\langle u,w\rangle|=\|u\|(|\lambda|\|u\|+|\gamma|\|v\|).$$
What can you conclude ($\|u\|\neq 0$)? You need to consider 2 different cases.
