Find minimum and maximum given $x+y+z = 10$ and $x^2+y^2+z^2 = 36 $. Given
$$ x+y+z = 10\\x^2+y^2+z^2 = 36 $$
find maximum and minimum value of $xyz$.
Using the relation A.M $\ge$ G.M it is possible to find maximum value as $ \frac{x+y+z}{3} \ge \sqrt[3]{xyz}$, but I have problem with finding minimum value, because harmonic mean isn't really helpful here. I think Lagrange multiplier may also be a solution but I'm curious if there's any easier option here.
 A: Using the first equation, $$x^2+y^2+z^2+2(xy+yz+zx)=100$$
From this equation, subtract the second equation, so we are left with,
$$2(xy+yz+zx)=64\\(xy+yz+zx)=32$$
Thus let x,y,z be the roots of the cubic eq P(a)=0 ,using vieta's theorem
$$a^3−10a^2+32a−p=0$$ where $p=xyz$.
Now you can find the maxima and minima of $a^3−10a^2+32a$ and hence  $p$. Apparently, the maxima and minima come out to be infinitely large. Why?
This is because, the domain of $a$ is such that we consider complex value of the variables named $(x,y,z)$. For the real values of the variables, it will only exist when $\dfrac{dp}{da}\leq 0$ (in other words we find the local extremas). Differentiating and equating to zero for $a^3−10a^2+32a$ gives us,
$$3a^2−20a+32=0$$ and the roots $a=\frac{8}{3}, 4$. Which further yields the maximum and minimum for the real domain as $32$ and $33.185$. You can verify this yourself following this.
A: HINT
For positive $x,y,z$
$$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+xz)$$
$$\implies (xy+yz+xz)=32$$
FOR MAXIMUM
Use AM $\ge GM$
$$xy+yz+xz \ge 3(xyz)^{\frac{2}{3}}$$
