<s>[INST] 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.
 [/INST] 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 $5,5,2$
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