Prove Cauchy-Schwarz equality. My professor asked me to prove the equality in Cauchy-Schwarz inequality. The equality holds iff the vectors $v$ and $u$ are linearly dependent.
I am able to show the equality using the fact $v$ and $u$ are linearly dependent.
But I don't know how to show the converse (i.e Showing the linear dependence using the equality).
 A: \begin{align}
\|\vec u+t\vec v\|^2 & = t^2 \|\vec v\|^2 + 2t\vec u\cdot\vec v + \|\vec u\|^2 \\[10pt]
& = at^2+bt+c.
\end{align}
This quadratic polynomial with $a>0$ is positive for every $t$ if the discriminant $b^2-4ac$ is negative, and is positive for all except one value of $t$ if the discriminant is $0$.  Express the discriminant in terms of $\|\vec u\|$, $\|\vec v\|$, and $\vec u\cdot\vec v$.
A: Hint: Let $z = pr_u (v) = \frac{\langle u,v\rangle}{\|u\|^2} u$  then we may write $v = z + w$, with $w \perp z$. By  Pythagoras Theorem $$\|v\|^2 = \|z\|^2 + \|w\|^2$$
Then $$\|v\| \geq \|z\|$$
And $\|v\| = \|z\| \implies\|w\| = 0$.
A: The CS inequality is: $|u\cdot v|^2 \leq|u|^2 |v|^2$ . 
Observe that : $\left(|u| - t|v|\right)^2 \geq 0, \forall t \in \mathbb{R}$ expanding this and treat it as a quadratic function in variable $t$. This is true for all $t$ implies $\triangle \leq 0 \to$ CS inequality follows, with equality when $|u| = t|v| \to u = \pm tv \to$ linear dependent.
A: Suppose that
$$
|u\cdot v|=\|u\|\,\|v\|\tag{1}
$$
This means that $u\cdot v=\pm\|u\|\,\|v\|$. Then
$$
\begin{align}
\|u-tv\|^2
&=(u-tv)\cdot(u-tv)\\
&=\|u\|^2\pm2t\|u\|\,\|v\|+t^2\|v\|\\
&=(\|u\|\pm t\|v\|)^2\tag{2}
\end{align}
$$
If $\|v\|=0$, then $u$ and $v$ are dependent (vacuously).
If $\|v\|\ne0$, then, by division, there is a $t$ so that $\|u\|\pm t\|v\|=0$. Using that $t$, $(2)$ says that $\|u-tv\|=0$, that is
$$
u=tv\tag{3}
$$
in which case $u$ and $v$ are dependent.
