Find the maximum and minimum value of $x$, where:
I thought I could use these values to form a equation having $x,y,z$ as roots and the sum of roots and $\sum{xy}$ but could not get any idea about the product of roots hence could not move ahead.
Let we set $x=\frac{4}{3}+u,y=\frac{4}{3}+v,z=\frac{4}{3}+w$. The given constraints translate into: $$\left\{\begin{array}{rcl}v+w &=& -u \\ v^2+w^2 &=& \frac{2}{3}-u^2\end{array}\right.$$ or: $$\left\{\begin{array}{rcl}v+w &=& -u \\ v\cdot w &=& u^2-\frac{1}{3},\end{array}\right.$$ so $v,w$ are roots of the polynomial $p(t)=t^2+ut+\left(u^2-\frac{1}{3}\right)$. They are real roots provided that $u^2-4\left(u^2-\frac{1}{3}\right)\geq 0$, so the critical values for $u$ are given by $u=\pm\frac{2}{3}$ and: $$ x\in\left[\frac{2}{3},2\right].$$
Hint for taking off:
Eliminate z between the two equations
$$ x^2 + y^2 -4 x + x y + 4 y + 5 = 0 $$
It is an ellipse. Intersection between a sphere and a plane is a circle, which projects to an ellipse.
Major axis inclined to axes due to presence of $ x y $ term.
Try differentiating with respect to $y$...
My Solution:: Given $$x+y+z=4$$ and $$x^2+y^2+z^2=6$$
So we get $$y+z=(4-x)$$ and $$y^2+z^2 = (6-x^2)$$
Now using the Cauchy-Schwarz inequality, we get
$$\displaystyle \left(y^2+z^2\right)\cdot (1^2+1^2)\geq (y+z)^2$$
So we get $$\displaystyle (6-x^2)\cdot 2 \geq (4-x)^2$$
$$\Rightarrow\displaystyle 12-2x^2\geq 16+x^2-8x$$
$$\Rightarrow\displaystyle 3x^2-8x+4\leq 0\Rightarrow 3x^2-6x-2x+4\leq 0$$
$$\Rightarrow\displaystyle 3x(x-2)-2(x-2)\leq 0\Rightarrow x\in \left[\frac{2}{3}\;,2\right]$$
