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.

You can also solve this problem using the method of Lagrange multipliers:
Note that $f(x,y,z) = xy + yz + xz$ is a $C^{\infty}$ function. We will use the method of Lagrange multipliers to determine the maximum/minimum of $f(x,y,z)$ given the restriction $g(x,y,z)= x^2 + y^2 + z^2 = 1$.
Therefore, because of Lagrange multipliers Theorem we know that if $x,y,z,\lambda \in R$ are solutions of the following system, then $(x,y,z)$ is either a max/min of $f(x,y,z)$ under the restriction $g(x,y,z) = 1 $. 
\begin{equation}
\begin{cases}
\nabla f(x,y,z) = \lambda \nabla g(x,y,z) , \quad  \lambda \neq 0 \Rightarrow \begin{cases} y + z = 2\lambda x \\
x + z = 2\lambda y \\
y+x = 2\lambda z
\end{cases}\\
g(x,y,z) = x^2 + y^2 + z^2 = 1
\end{cases}
\end{equation}
Solving the system we get the following values for $(x,y,z,\lambda)$:
\begin{equation}
(x,y,z,\lambda) =\left(x, \frac{1}{2} (-\sqrt{2 - 3 x^2} - x), \frac{1}{2} (\sqrt{2 - 3 x^2} - x), -\frac{1}{2}\right)
\end{equation} 
\begin{equation}
(x,y,z,\lambda) = \left(x, \frac{1}{2} (-\sqrt{2 - 3 x^2} - x), \frac{1}{2} (\sqrt{2 - 3 x^2} - x), -\frac{1}{2} \right)
\end{equation}
\begin{equation}
(x,y,z,\lambda) = \left( \frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}} ,\frac{1}{\sqrt{3}} ,1 \right)
\end{equation}
\begin{equation}
(x,y,z,\lambda) = \left( -\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}} ,-\frac{1}{\sqrt{3}} ,1 \right)
\end{equation}
In order to know whether the points obtained solving the system are max or min, we look evaluate them:
\begin{equation}
f\left(x, \frac{1}{2} (-\sqrt{2 - 3 x^2} - x), \frac{1}{2} (\sqrt{2 - 3 x^2} - x)\right) = -\frac{1}{2}
\end{equation}
\begin{equation}
f\left(x, \frac{1}{2} (\sqrt{2 - 3 x^2} - x), \frac{1}{2} (-\sqrt{2 - 3 x^2} - x)\right) = -\frac{1}{2}
\end{equation}
\begin{equation}
f \left( \frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}} ,\frac{1}{\sqrt{3}}\right) = 1
\end{equation}
\begin{equation}
f \left( -\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}} ,-\frac{1}{\sqrt{3}}\right) = 1
\end{equation}
Therefore we have two minimums at $\left(x, \frac{1}{2} (-\sqrt{2 - 3 x^2} - x), \frac{1}{2} (\sqrt{2 - 3 x^2} - x)\right)$ and $\left(x, \frac{1}{2} (\sqrt{2 - 3 x^2} - x), \frac{1}{2} (-\sqrt{2 - 3 x^2} - x)\right)$ $\forall x\in R$, and a maximum at $\left( \frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}} ,\frac{1}{\sqrt{3}}\right)$ and $\left( -\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}\right)$. 
Now, to prove that the minimum is indeed global you can perfectly use your steps and conclude that $f(x,y,z) \geq -\frac{1}{2}$ and so, the minimum is $-\frac{1}{2}$. 
In fact, proving that $f(x,y,z) \geq -\frac{1}{2}$ and then finding some $x_0,y_0,z_0 \in \mathbb{R}$ for which $f(x_0,y_0,z_0) = -\frac{1}{2}$ is sufficient and necessary to state that $-\frac{1}{2}$ is the minimum. However, by using Lagrange multipliers you will certainly find those $x_0,y_0,z_0$ without "making them up".
