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.

Here's another way to approach this. It's easy to see that the value of $\frac13$ is obtained when each of $x, y, z$ is $\frac13$. We want to show that as the variables deviate from this point (with their sum still being 1) the value cannot decrease.
So we look at the deviations from $\frac13$: $x=\frac13+\epsilon_1$, $y=\frac13+\epsilon_2$, $z=\frac13+\epsilon_3$ with $\epsilon_1+\epsilon_2+\epsilon_3=0$. you have
$x^2+y^2+z^2=\\
(\frac13+\epsilon_1)^2+(\frac13+\epsilon_2)^2+(\frac13+\epsilon_3)^2=\\\left(\frac19+\frac23\epsilon_1+\epsilon_1^2\right)+\left(\frac19+\frac23\epsilon_2+\epsilon_2^2\right)+\left(\frac19+\frac23\epsilon_3+\epsilon_3^2\right)=\\
\left(\frac19+\frac19+\frac19\right)+\frac23(\epsilon_1+\epsilon_2+\epsilon_3)+(\epsilon_1^2+\epsilon_2^2+\epsilon_3^2)=\\
\frac13+(\epsilon_1^2+\epsilon_2^2+\epsilon_3^2) \ge \frac13$
