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
 A: 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 $$\frac{10}{3}-(-2)=\frac{16}{3}.$$
A: We have $$
a+b=(a,b)\cdot (1,1)\leq (a^2+b^2)^{1/2}\sqrt2, $$ which they use as $$(a+b)^2\leq 2 (a^2+b^2). $$
A: 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 $c=10/3$.

A: 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 $\dfrac{16}{3}$.
A: 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.
A: 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, $$-2\leq c \leq \frac{10}{3}.$$
