Find the value of $xyz$ 
$x$, $y$, $z$ are real numbers such that
$①$  $ x+y+z=\sqrt{3}$
$②$ $xy+yz+zx=1$
In this case, find the value of $xyz$.

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Here's my attempt.
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$3=(x+y+z)^2 = x^2 + y^2 + z^2 +  2(xy+yz+zx) = x^2 + y^2 + z^2  + 2$
$\therefore   \ \  x^2 + y^2 + z^2 = 1 $
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$   x^3  +y^3 +  z^3    -   3xyz=(x+y+z)(x^2 +  y^2  +  z^2    -  xy-yz-zx)  = 0$
$\therefore   \ \  x^3 + y^3 + z^3 = 3xyz $
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What can I do... after these?
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I guessed the answer like this.
If $ x=y=z = \frac{\sqrt{3}}{3}$ , then these satisfy $①, ②$.
So, $\  $ $xyz=\frac{\sqrt{3}}{9}$....
Thanks in advance.

Another property of AM-GM is that if the arithmetic means and geometric means are equal, all the variables are equal.
Since $\frac{x^3+y^3+z^3}{3} ≥ ({x^3 \cdot y^3 \cdot z^3})^{1/3}$, we get  $x^3+y^3+z^3 ≥ xyz$, and as a consequence $x = y = z$. Therefore, you only need to solve the equation $u^3 + u^3 + u^3 = u \cdot u \cdot u$, and then substitute into equation $1$.
