# Solve this without Cauchy-Schwartz?

$a+b+c=2$
$a^2+b^2+c^2=12$.

($a,b,c$) are not necessarily integers. Find the difference between the maximum possible and the minimum possible value of $c$.

I have seen a solution to the problem using the Cauchy-Schwarz inequality, but is there any other way to do this? And please keep it to a level where I can understand? Thanks.

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A very bad solution: The solution set of the system is the intersection of a plane and a sphere, hence is a circle. So you can parametrize it using only one parameter t. Then study c(t). –  Taladris May 6 '13 at 2:09

another way is:

$(a+b)^2=a^2+b^2+2ab \le a^2+b^2+a^2+b^2 \to a^2+b^2 \ge \dfrac{(a+b)^2}{2}$

$12=a^2+b^2+c^2 \ge \dfrac{(a+b)^2}{2}+c^2=\dfrac{(2-c)^2}{2}+c^2 \to 3c^2-4c+4 \le 24 \to 3c^2-4c-20 \le 0 \to (3c-10)(c+2) \le 0 \to -2 \le c \le \dfrac{10}{3}$

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Thank you very much! –  Ovi May 6 '13 at 1:33
We are looking at where a plane that has symmetry in the $3$ variables meets a sphere.
The highest and lowest points have equal $x$ and $y$ coordinates. So we get the equations $2a+c=2$, $2a^2+c^2=12$. Eliminate $a$. We have $4a^2=(2-c)^2$ and $4a^2=24-2c^2$, giving $3c^2-4c-20=0$. Now we can calculate the two values of $c$. They even happen to be rational.
It is just a question of tilting one's head to see the geometry. If one is more algebraically minded, use Lagrange multipliers. We get $\lambda +2\mu a=0$, $\lambda+2\mu b=0$, $1+\lambda+2\mu c=0$, giving $a=b$. –  André Nicolas May 6 '13 at 1:43