<s>[INST] Numbers $x,y,z$ are such that $x+y+z=0$ and $x^2+y^2+z^2=1$. Show that at least one of $xy,xz,yz$ is not bigger than $-\frac{1}{3}$. So i had to prove:

Numbers $x,y,z$ are such that $x+y+z=0$ and $x^2+y^2+z^2=1$. Show that at least one of $xy,xz,yz$ is not bigger than $-\frac{1}{3}$.

Here is my proof:
So we have
 $(x+y+z)^2=x^2+y^2+z^2+2xy+2yz+2zx=1+2(xy+yz+zx)=0\Rightarrow2(xy+yz+xz)=-1$
Let's assume that all of $xy,zx,yz$ are bigger than $-\frac{1}{3}$
$xy>-\frac{1}{3}$
$yz>-\frac{1}{3}$
$zx>-\frac{1}{3}$ 
Adding that we have $xy+yz+zx>-1$
$2(xy+yz+zx)>xy+yz+zx>-1\Rightarrow 2(xy+yz+zx)>-1$
Which is contradiction because $2(xy+yz+xz)=-1$, so one of $xy,yz,zx$ must be less or equal $-\frac{1}{3}$
Is my proof valid?
 [/INST] I found another, easier answer, so I thought it might be worth posting it.
Without loss of generality we may assume that $x \leqslant y \leqslant z$. From our assumptions we see that $x<0$ and $0<z<1$. If $0 \,\leqslant y$, then
$$1=x^2+y^2+z^2=(-y-z)^2+y^2+z^2=2y^2+2z^2+2yz \geqslant 2y^2+2y^2+2y^2=6y^2$$
If $y<0%$:
$$1=x^2+y^2+z^2=x^2+y^2+(-x-y)^2=2x^2+2y^2+2xy \geqslant 2y^2+2y^2+2y^2=6y^2$$
In both cases we have: $\; y^2\leqslant \frac16\ $, and also $1=x^2+(-x-z)^2+z^2=2x^2+2z^2+2xz \,$,which means that:
$$xz= \frac12-(x^2+z^2)=\frac12-(1-y^2)=y^2-\frac12\leqslant\frac16-\frac12=-\frac13 $$
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