Maximum minus minimum of $c$ where $a+b+c=2$ and $a^2+b^2+c^2=12$ Let $a,b,$ and $c$ be real numbers such that
$a+b+c=2  \text{ and } a^2+b^2+c^2=12.$
What is the difference between the maximum and minimum possible values of $c$?
$\text{(A) }2\qquad \text{ (B) }\frac{10}{3}\qquad \text{ (C) }4 \qquad \text{ (D) }\frac{16}{3}\qquad \text{ (E) }\frac{20}{3}$
As I was reading the solution for this problem, I noticed that it said to use Cauchy–Schwarz inequality. I know what this inequality is (dot product of two vectors < vector 1*vector 2), but I don't understand how it can be applied in this situation. Thanks!
https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12B_Problems/Problem_17

Let $a=mc$ and $b=nc$ then 
$$m+n=\dfrac{2}{c}-1~~~,~~~m^2+n^2=\dfrac{12}{c^2}-1$$
by Cauchy-Schwarz 
$$(m+n)^2\leq2(m^2+n^2)$$
with substitution $-3c^2+4c+20\geq0$ gives $c=-2,\dfrac{10}{3}$ leads us to difference $\dfrac{16}{3}$.
