Prove: $|\langle u,v \rangle|= \|u\|\cdot \|v\|\iff u=\alpha v$ 
Let $V$ be an inner product space s.t $u,v\in V$ and $u,v\neq 0$
Prove:  $$\langle u,v \rangle = ||u||\cdot ||v||\iff u=\alpha v$$ s.t $\alpha\in \mathbb{C}$

$\Leftarrow:$
$$|\alpha| \langle  v,v \rangle=|\alpha\langle  v,v \rangle|=|\langle  \alpha v,v \rangle|=|\langle u,v \rangle|=|\langle u,v \rangle| \leq |\|u\|\cdot\|v\| |=| \|\alpha v\|\cdot\|v\| |=|\alpha |||v||^2=|\alpha| \langle  v,v \rangle$$
I am left with $|\alpha|$
How should I approach the proof of $\Rightarrow$
 A: The statement is incorrect.

You can consider the orthogonal projection of $u$ on the subspace generated by $v$: then $u=\alpha v+w$, where $\langle w,v\rangle=0$.
Now, assuming linearity in the first variable,
$$
\langle u,v\rangle=\langle \alpha v+w,v\rangle=\alpha\langle v,v\rangle
$$
Also
$$
\|u\|^2=\langle \alpha v+w,\alpha v+w\rangle=|\alpha|^2\langle v,v\rangle+\langle w,w\rangle
$$
Assuming $\|u\|\cdot\|v\|=\langle u,v\rangle$, we obtain
$$
\alpha^2\langle v,v\rangle^2=
(|\alpha|^2\langle v,v\rangle+\langle w,w\rangle)\langle v,v\rangle
$$
that simplifies to
$$
(\alpha^2-|\alpha|^2)\|v\|=\|w\|
$$
In particular $\alpha^2$ must be real and nonnegative, which forces $\alpha$ to be real, so $\alpha^2=|\alpha|^2$ and therefore $\|w\|=0$.
It would be similar with linearity in the second variable.

For the $\Leftarrow$ direction, if $u=\alpha v$, then
$$
\langle u,v\rangle=\alpha\|v\|
$$
and $\|u\|\,\|v\|=|\alpha|\,\|v\|$, so we get the equality only if $\alpha$ is real and nonnegative.

A correct statement would be:

Let $V$ be a real inner product space, $u,v\in V$ be nonzero vectors. Then $\langle u,v\rangle=\|u\|\,\|v\|$ if and only if $u=\alpha v$ for a positive scalar $\alpha$.

A: Notice that for $y \ne 0$ we have:
$\newcommand\inner[2]{\left\langle #1, #2 \right\rangle}$
$$\inner{x}{x}\inner{y}{y} - \left|\inner{x}{y}\right|^2 = \frac{1}{\inner{y}{y}}\Big\langle\inner{y}{y}x - \inner{x}{y}y, \inner{y}{y}x - \inner{x}{y}y\Big\rangle$$
Indeed:
\begin{align}
\Big\langle\inner{y}{y}x - \inner{x}{y}y, \inner{y}{y}x - \inner{x}{y}y\Big\rangle &= \inner{y}{y}^2 \inner{x}{x} - \inner{y}{y}\overline{\inner{x}{y}}\inner{x}{y} - \inner{x}{y}\inner{y}{y}\inner{y}{x} + \left|\inner{x}{y}\right|^2 \inner{y}{y} \\
&= \inner{y}{y}^2 \inner{x}{x} - \left|\inner{x}{y}\right|^2\inner{y}{y}\\
&= \inner{y}{y}\Big(\inner{y}{y} \inner{x}{x} - \left|\inner{x}{y}\right|^2\Big)
\end{align}
If equality in CSB holds, we have $\inner{y}{y} \inner{x}{x} - \left|\inner{x}{y}\right|^2 = 0$, so using definiteness of the inner product we obtain $\inner{y}{y}x - \inner{x}{y}y = 0$. Thus, $x = \frac{\inner{x}{y}}{\inner{y}{y}}y$.
On the other hand, if $y = 0$, the equality holds, and $x$ is obviously proportional to $y$.
A: If $\|v\|\ne0$, then
$$
\begin{align}
\left\|u-v\frac{\langle u,v\rangle}{\|v\|^2}\right\|^2
&=\left\langle u-v\frac{\langle u,v\rangle}{\|v\|^2},u-v\frac{\langle u,v\rangle}{\|v\|^2}\right\rangle\tag1\\
&=\|u\|^2-2\frac{\langle u,v\rangle^2}{\|v\|^2}+\|v\|^2\frac{\langle u,v\rangle^2}{\|v\|^4}\tag2\\
&=\|u\|^2-\frac{\langle u,v\rangle^2}{\|v\|^2}\tag3\\
&=\|u\|^2-\frac{\|u\|^2\|v\|^2}{\|v\|^2}\tag4\\[6pt]
&=0\tag5
\end{align}
$$
Explanation:
$(1)$: $\|u\|^2=\langle u,u\rangle$
$(2)$: expand
$(3)$; simplify
$(4)$: apply hypothesis that $|\langle u,v\rangle|=\|u\|\,\|v\|$
$(5)$: simplify
Thus,
$$
u=v\,\overbrace{\ \ \frac{\langle u,v\rangle}{\|v\|^2}\ \ }^\alpha
$$
A: Suppose that $\langle u,v\rangle=\|u\|.\|v\|$. Consider the map$$\begin{array}{rccc}f\colon&\mathbb{R}&\longrightarrow&\mathbb R\\&\lambda&\mapsto&\|u+\lambda v\|^2.\end{array}$$You know that $(\forall\lambda\in\mathbb{R}):f(\lambda)\geqslant0$. But$$f(\lambda)=\|u\|^2+2\lambda\langle u,v\rangle+\lambda^2\|v\|^2.$$If $v=0$, then $u=0.v$. Otherwise, $f$ is a quadratic function. Its discriminant is $4\langle u,v\rangle^2-4\|u\|.\|v\|=0$. Since it is $0$, $f$ must have some root. Let $\alpha$ be the symmetric of such root. Then $f(-\alpha)=0$. In other words, $\|u-\alpha v\|=0$ and so $u=\alpha v$.
A: Two vectors $x$ and $y$ are linearly dependent if and only if
$$\langle y,y\rangle x = \langle y,x \rangle y$$
Proof:
Suppose
$$ax+by=0$$
for some $(a,b)\ne (0,0)$.
Then WLOG let $a \ne 0$, and we have
$$x=-\frac{b}{a}y$$
and hence
$$\langle y,y \rangle x - \langle y,x \rangle y = \langle y,y \rangle \frac{-b}{a}y - \langle y, -\frac{b}{a} y \rangle y = -\frac{b}{a}\langle y,y \rangle y +\frac{b}{a} \langle y,y \rangle y = 0$$
Conversely, if
$$\langle y,y \rangle x - \langle y,x \rangle y = 0$$
then either $\langle y,y \rangle =0$ or $x$, $y$ are linearly dependent.
But
$$\langle y,y \rangle =0 \iff y =0$$
and hence $x$, $y$ are linearly dependent.
A: for $x\in \Bbb R$, set the quadratic
 $$P(x)  = \|u+x v\|^2 =\|u\|^2+2x\langle u,v\rangle+x^2\|v\|^2.$$
then, 
$$|\langle u,v \rangle |= \|u\|\cdot \|v\|\Longleftrightarrow  \Delta = 4|\langle u,v\rangle|^2 -4\|u\|^2\|v\|^2 = 0\\~~~~~~~~~~~~~\Longleftrightarrow  x_0=-\frac{b}{2a}=\frac{-\langle u,v\rangle}{\|v\|^2}~\\~~~\text{is the solution to $P(x)=0.$}$$
but $$ 0 = P(x_0) = \|u+x_0 v\|^2 \implies  u =-x_0v $$
