<s>[INST] 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
 [/INST] We wish to find the values of $c$ such that the following simultaneous equations have real solutions to $a$ and $b$:
$$\begin{cases}a+b&=&2-c\\a^2+b^2&=&12-c^2\end{cases}$$
Labelling the equations as $(1)$ and $(2)$ respectively, $(1)^2 - (2)$ gives $2ab 
= -8-4c+2c^2$, so $(a-b)^2 = a^2+b^2 - 2ab = (12-c^2) - (-8-4c+2c^2) = 20+4c-3c^2$.
If that number is non-negative, then $a-b = \pm\sqrt{20+4c-3c^2}$, which together with $(1)$ gives two sets of solutions.
So it remains to solve $20 + 4c - 3c^2 \ge 0$, no Cauchy-Schwarz needed.
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