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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

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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 $c=10/3$.

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  • $\begingroup$ how is that using cauchy though? I don't understand how we can take these variables and pull out vectors. Sorry if this question is very elementary lol $\endgroup$ – Dude156 Jul 31 '18 at 5:10
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    $\begingroup$ @Dude156 Consider the vectors $v=(a,b)$ and $w=(1,1)$. Then $v \cdot w = a + b$ while $\|v\| \|w\| = \sqrt{(a^2 + b^2)(1^2 + 1^2)}$. $\endgroup$ – angryavian Jul 31 '18 at 5:14
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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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We have $$ a+b=(a,b)\cdot (1,1)\leq (a^2+b^2)^{1/2}\sqrt2, $$ which they use as $$(a+b)^2\leq 2 (a^2+b^2). $$

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  • $\begingroup$ $c^2\leq3 (16-c^2),$ implies $c^2\leq12$ $\endgroup$ – Nosrati Jul 31 '18 at 4:57
  • $\begingroup$ Of course. Thanks, I'll edit right away. $\endgroup$ – Martin Argerami Jul 31 '18 at 4:59
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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 $\dfrac{16}{3}$.

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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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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, $$-2\leq c \leq \frac{10}{3}.$$

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