Range of $xyz\;,$ If $x+y+z=4$ and $x^2+y^2+z^2=6$ 
If $x,y,z\in \mathbb{R}$ and $x+y+z=4$ and $x^2+y^2+z^2=6\;,$ Then range of $xyz$

$\bf{My\; Try::}$Using $$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)$$
So we get $$16=6+2(xy+yz+zx)\Rightarrow xy+yz+zx = -5$$ and given $x+y+z=4$
Now let $xyz=c\;,$ Now leyt $t=x,y,z$ be the roots of cubic equation, Then
$$\displaystyle (t-x)(t-y)(t-z)=0\Rightarrow t^3-(x+y+z)t^2+(xy+yz+zx)t-xyz = 0$$
So we get $\displaystyle t^3-4t^2-5t-c=0$
Now let $f(t)=t^3-4t^2-5t-c\;,$ Then $f'(t)=3t^2-8t-5$
and $f''(t)=6t-8.$ Now for max. and Min.$f'(t)=0\Rightarrow 3t^2-8t-5=0$
So we get $\displaystyle t=\frac{8\pm \sqrt{64+60}}{2\cdot 3}=$
Now How can I solve it after that, Help required, Thanks

Note that $$c=t((t-2)^2+1),\ t\in \mathbb{R}$$, so according to the calculations, $c\in [a,b]$ where $$a=t((t-2)^2+1)|_{t=5/3}=\frac{50}{27},\ b=t((t-2)^2+1)|_{t=1}=2$$
