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

Solution
Notice that $$a+b=2-c\tag1,$$and $$a^2+b^2=12-c^2.\tag2$$
Since $$ 2ab \leq a^2+b^2,$$Hence$$(a+b)^2=a^2+b^2+2ab \leq 2(a^2+b^2).\tag3$$Put $(1),(2)$ into $(3)$. We obtain $$(2-c)^2 \leq 2(12-c^2),$$namely $$3c^2-4c-20=(c+2)(3c-10) \leq 0.$$
As a result, $$-2\leq c \leq \frac{10}{3}.$$
