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

Hint:
$$(2-c)^2 = (a+b)^2 = (a \cdot 1 + b \cdot 1)^2 \le (a^2 + b^2) (1^2 + 1^2) = 2(12-c^2)$$
Finding $c$ satisfying this inequality amounts to solving a quadratic.

 $$c^2 - 4c + 4 \le 24 - 2 c^2$$ $$3 c^2 - 4 c - 20 \le 0$$ So $c$ is between the two roots $-2$ and $10/3$. Note that you should check that for each of these values of $c$, there exist valid choices of $a$ and $b$ that satisfy the two original equalities. Specifically, $a=b=2$ and $c=-2$ works, as well as $a=b=-2/3$ and $c=10/3$.

