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?
 A: Regarding my notations here, I refer to my comments under the question.  The idea of the parametrization in my comments arises from the following observation.  
For $(x,y,z)\in S$, the maximum value of $|x|$ is $\sqrt{2/3}$.  To show this, we note that $x=-y-z$ and
$$1=x^2+y^2+z^2=(-y-z)^2+y^2+z^2=2(y^2+yz+z^2) \geq \frac{3}{2}\left(y+z\right)^2=\frac{3}{2}x^2,$$
since $$4(y^2+yz+z^2)=(y^2+z^2)+(3y^2+4yz+3z^2)\geq 2yz+(3y^2+4yz+3z^2)=3(y+z)^2.$$
The maximum value $|x|=\sqrt{\frac23}=\frac2{\sqrt6}$ is attained when $(x,y,z)=\pm\left(\frac{2}{\sqrt6},-\frac{1}{\sqrt6},-\frac{1}{\sqrt6}\right)$.  Hence we may write $x=\sqrt{\frac{2}{3}}\cos t$ for some $t\in[0,2\pi)$.  Due to symmetry, we expect $y=\sqrt{\frac{2}{3}}\cos\left(t-\frac{2\pi}{3}\right)$ and $z=\sqrt{\frac{2}{3}}\cos\left(t+\frac{2\pi}{3}\right)$.  We now need to check whether $x+y+z=0$ and $x^2+y^2+z^2=1$ hold, but this is an easy exercise.  (Note that for a given $x$, $(y,z)$ is unique up to swapping because $q=y$ and $q=z$ are roots of $q^2+xq+x^2-\frac12$.  Furthermore, the assignment $t\mapsto 2\pi-t$ swaps the values of $y$ and $z$ while keeping $x$ fixed.  Thus, this parametrization gives all points on $S$.)
Now, as mentioned by YiFan, the function $f:S\to\Bbb R$ defined by $$f(p)=\min\{xy,yz,zx\}$$ for $p=(x,y,z)\in S$ is invariant under permuting entries of $p$ and also under the antipodal map $p\mapsto -p$.  Therefore we can assume that $x>0\geq y\geq z$, which ultimately implies that $t\in[0,\pi/6]$.  For such points $p$, $$f(p)=xz=\frac23\cos t\cos\left(t+\frac{2\pi}{3}\right)=\frac{1}{3}\Biggl(\cos\frac{2\pi}{3}+\cos\left(2t+\frac{2\pi}{3}\right)\Biggr).$$
For $t\in[0,\pi/6]$, the maximum value of $\cos\left(2t+\frac{2\pi}{3}\right)$ is attained at $t=0$ since $$\frac{2\pi}{3}\leq  2t+\frac{2\pi}{3}\leq \pi.$$  Therefore the maximum value of $f(p)$ is 
$$\frac{1}{3}\Biggl(\cos\frac{2\pi}{3}+\cos\left(2\cdot 0+\frac{2\pi}{3}\right)\Biggr)=\frac{1}{3}\left(-\frac12-\frac12\right)=-\frac13.$$
(It is also not difficult to see that the minimum value of $f(p)$ is $-1/2$.  This is attained when $p$ is a permutation of $\left(-\frac1{\sqrt2},0,\frac1{\sqrt2}\right)$; that is $-\frac12\le\min\{xy,yz,zx\}\le-\frac13$.  We can also show that $0\le \max\{xy,yz,zx\}\le\frac16$ and $-\frac13\le\operatorname{median}\{xy,yz,zx\}\le0$.)
Alternatively if $\mu=\min\{xy,yz,zx\}$, then show that
$$\operatorname{median}\{xy,yz,zx\}+\max\{xy,yz,zx\}=-\frac{1+2\mu}{2}$$
and
$$\operatorname{median}\{xy,yz,zx\}\cdot\max\{xy,yz,zx\}=\frac{\mu(1+2\mu)}{2}.$$  This implies
$$\operatorname{median}\{xy,yz,zx\}=\frac{-(1+2\mu)-\sqrt{(1-6\mu)(1+2\mu)}}{4}$$
and
$$\max\{xy,yz,zx\}=\frac{-(1+2\mu)+\sqrt{(1-6\mu)(1+2\mu)}}{4}.$$
This means $\mu\geq -1/2$.  We must have
$$\mu \leq \frac{-(1+2\mu)-\sqrt{(1-6\mu)(1+2\mu)}}{4},$$
or
$$\sqrt{(1-6\mu)(1+2\mu)}\leq -(1+6\mu).$$
That is,
$$1-4\mu-12\mu^2\leq 1+12\mu+36\mu^2,$$
or
$$16\mu(1+3\mu)\ge0.$$
It is clear that $\mu<0$, so $1+3\mu \leq 0$, and hence
$$\mu\le -\frac13.$$
A: 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:
$$xz= \frac12-(x^2+z^2)=\frac12-(1-y^2)=y^2-\frac12\leqslant\frac16-\frac12=-\frac13 $$
