Knowing that for any set of real numbers $x,y,z$, such that $x+y+z = 1$ the inequality $x^2+y^2+z^2 \ge \frac{1}{3}$ holds. 
Knowing that for any set of real numbers $x,y,z$, such that $x+y+z = 1$ the inequality $x^2+y^2+z^2 \ge \frac{1}{3}$ holds.

I spent a lot of time trying to solve this and, having consulted some books, I came to this: 
$$2x^2+2y^2+2z^2 \ge 2xy + 2xz + 2yz$$ $$2xy+2yz+2xz = 1-(x^2+y^2+z^2) $$ $$2x^2+2y^2+2z^2 \ge 1 - x^2 -y^2 - z^2 $$ $$x^2+y^2+z^2 \ge \frac{1}{3}$$
But this method is very unintuitive to me and I don't think this is the best way to solve this. Any remarks and hints will be most appreciated.

We have an instance of the least-norm problem
$$\begin{array}{ll} \text{minimize} & \| {\rm x} \|_2^2 \\ \text{subject to} & {\Bbb 1}_3^\top {\rm x} = 1\end{array}$$
The minimizer is
$${\rm x}_{\min} := {\Bbb 1}_3 \left( {\Bbb 1}_3^\top {\Bbb 1}_3 \right)^{-1} = \frac13 {\Bbb 1}_3$$
and, thus, the minimum is $\| {\rm x}_{\min} \|_2^2 = \frac13$. Hence, $\color{blue}{  \| {\rm x} \|_2^2 \geq \frac13 }$.
