Given $x,y,z\geq0$ and $x^2+y^2+z^2+x+2y+3z=13/4$. Find the minimum of $x+y+z$. 
Given $x,y,z\geq0$ and $x^2+y^2+z^2+x+2y+3z=13/4$. Find the minimum of $x+y+z$.

I tried many method, such as AM-GM, but all of them failed.
Thank you.
 A: use this 
$$(x+y+z)^2+3(x+y+z)\ge x^2+y^2+z^2+x+2y+3z$$
and let $x+y+z=t$,
then $$t^2+3t\ge \dfrac{13}{4}$$
A: Geometrically the given equation represents a sphere and $x+y+z=t$ represents a plane which can be shifting around. So the min will be achieved on $x,y$ or $z$-axes intersecting with the sphere. Hence the possible points are $(\frac{\sqrt{14}-1}{2},0,0),(0,\frac{\sqrt{17}}{2}-1,0),(0,0,\frac{\sqrt{22}-3}{2})$, and the min is $\frac{\sqrt{22}-3}{2}$.
A: The equation in fact is the surface of a ball at center $C_0=(-1/2,-1,-3/2)$ with radius $r=3\sqrt{3}/2$.
Then consider when there is only one point on the intersection between sphere and plane $x+y+z=t$
A: You can rewrite the equation as $$(x+\frac12)^2+(y+1)^2+(z+\frac32)^2=\frac{27}4$$
As others have mentioned, this is the equation of a sphere. When $x=y=z=0$ the left hand side of this equation is $\dfrac72$, and this is too low. Increasing $x$ or $y$ or $z$ increases the value, and to keep $x+y+z$ as small as possible you need to increase the value as fast as possible. Thinking about the square function, the gradient gets steeper as you move away from $0$, so you want to be as far out as possible. $z+\dfrac32$ starts larger than the other two squares, so you increase fastest by setting $x=y=0$ and $z$ as large as possible.
This comes out as $$\frac14+1+(z+\frac32)^2=\frac{27}4$$
$$z+\frac32=\sqrt{\frac{11}2}$$
