Prove that one of x,y,z is smaller than 3 and one is bigger than 5 if... If $x+y+z=12$ and $x^2+y^2+z^2=54$ then prove that one has to be smaller or equal to 3 and one has to be bigger or equal than 5.
So I got that $xy+yz+zx=45$ and with that I had a function with x,y,z as zeros: $f(a)=a^3-12a^2+45a-k$, where k=xyz.
I get that $f'(a)=3(a-3)(a-5)$, but I can't prove what I need with that.
 A: Fleshing out the comment by peterwhy.
Assume first that $x,y,z$ are all distinct. W.l.o.g. we can assume that $x>y>z$. Then $x,y,z$ are the three zeros of $f(a)$. By Rolle's theorem $f'(a)$ must have a zero between $z$ and $y$ and another zero between $y$ and $x$. But you know the zeros of $f'(a)$, so this shows that
$$z<3<y<5<x.$$
Ok. That leaves us with the case when two or more of them are equal. Without loss of generality we assume that $x=y$. Then you get the system $2x+z=12$, $2x^2+z^2=54$. I'm sure you can find the solutions of that system, and verify the claim in this case also.
A: Here is what I got:
Let $x\ge y\ge z> 3$, then we that we can write x,y,z as follow: $a=x+3,b=y+3,c=z+3$. Then we have that $a+b+c=3$ and $a^2+b^2+c^2=9$. With that we have that $ab+bc+ca=0$, but bcs $x,y,z>3>0$ then are a,b,c positive too, so it's impossible that ab+bc+ca=0, contradiction. One has to be smaller than 3.
The same thing goes for proving $x\ge 5$, we say that $5>x\ge y\ge z$ and write the same thing and get again that ab+bc+ca=0.
A: A sketch that needs fleshing out:  We might as well require $x \ge y \ge z$. If $z \gt 3$, let $w=3-z$ and we are thinking of $w$ as small.  The largest $x^2+y^2+z^2$ can be is $(6-2w)^2+2(3+w)^2=54-12w+6w^2$, so $z \lt 3$  If $x \lt 5$, Let $x=5-w$, then the largest $x^2+y^2+z^2$ can be is $2(5-w)^2+(2+2w)^2=54-16w+6w^2$.  In fact there is a solution $5,5,2$
