<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] By C-S
$$12=a^2+b^2+c^2=\frac{1}{2}(1^2+1^2)(a^2+b^2)+c^2\geq\frac{1}{2}(a+b)^2+c^2=\frac{1}{2}(2-c)^2+c^2,$$
which gives $$3c^2-4c-20\leq0$$ or
$$(3c-10)(c+2)\leq0$$ or
$$-2\leq c\leq\frac{10}{3}.$$
Now, we get $$\frac{10}{3}-(-2)=\frac{16}{3}.$$
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