<s>[INST] Find the minimum value of $xy+yz+xz$ Question:

Find the minimum value of $xy+yz+xz$, given that $x,y,z$ are real and $x^2+y^2+z^2=1$

My attempt, 
Since $(x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz)$
$=1+2(xy+xz+yz)$
$(x+y+z)^2\geq0$
So that $1+2(xy+xz+yz)\geq0$
$(xy+xz+yz)\geq -\frac{1}{2}$
So the minimum value is $-\frac{1}{2}$.
My question is am I correct ? Because I don't have the solution. And is there another way to solve this? Thanks in advance.
 [/INST] You are almost done. You are right that, since $(x+y+z)^2 \geq 0$, we have $xy+xz+yz \geq -\frac{1}{2}$. Now, to find the actual minimum, turn the inequality sign into an equality sign:
$$x+y+z = 0 \Rightarrow xy+xz+yz=-\frac{1}{2}$$
Now, just find values of $x, y, z$ such that
$$\begin{cases} x^2+y^2+z^2=1 \\ x+y+z=0 \end{cases}$$
This is an equation of a circle (sphere intersected with a plane), and all it's points give you the minimum. To find one, you can simply take almost any values for $x,y,z$, then project this point on the plane, and then normalize the vector so that it lies on the sphere.
For example, take $x=0, y=0, z=-1$. One possible way of projecting on the plane $x+y+z=0$ is to take the sum of coordinates $s$, and decrease every coordinate by $\frac{s}{3}$. For the chosen vector we get $s=-1$, and the vector transforms into $$x=\frac{1}{3}, y=\frac{1}{3}, z=-\frac{2}{3}$$
Now, compute this vector's norm: $$\sqrt{x^2+y^2+z^2} = \sqrt{\frac{1^2+1^2+2^2}{3^2}} = \sqrt{\frac{6}{9}} = \sqrt{\frac{2}{3}}$$
Now, dividing the vector's coordinates by the norm (thus multiplying by $\sqrt{\frac{3}{2}}$), we get a vector on the sphere (and, since it remains being in the plane, it appears to be on the desired circle):
$$x = \frac{1}{3}\sqrt{\frac{3}{2}}=\frac{1}{\sqrt{6}}$$
$$y = \frac{1}{3}\sqrt{\frac{3}{2}}=\frac{1}{\sqrt{6}}$$
$$z = -\frac{2}{3}\sqrt{\frac{3}{2}}=-\frac{2}{\sqrt{6}}$$
Note that it is exactly the solution by CY Kwong in the comments.
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