How to find minimum and maximum value of x, if x+y+z=4 and $x^2 + y^2 + z^2 = 6$? I just know that putting y=z, we will get 2 values of x. One will be the minimum and one will be the maximum. What is the logic behind it?

The second condition gives
$$x^2+(y+z)^2-2yz=6$$ or
$$x^2+(4-x)^2-2yz=6$$ or
$$yz=x^2-4x+5,$$
which with $y+z=4-x$ gives
$$(4-x)^2-4(x^2-4x+5)\geq0$$ or
$$3x^2-8x+4\leq0$$ or
$$\frac{2}{3}\leq x\leq2.$$
The equality occurs for $y=z$ because $y$ and $z$ are roots of the equation
$$t^2-(4-x)t+x^2-4x+5=0$$ and the equality in $\Delta\geq0$ happens for $y=z$.
Id est, $$\max_{x+y+z=4,x^2+y^2+z^2=6}x=2$$ and
 $$\min_{x+y+z=4,x^2+y^2+z^2=6}x=\frac{2}{3}.$$ 
Done!
