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.

One more way to prove it is by substituting out for $z$:
$$
x^2+y^2+z^2 = x^2+y^2+(1-x-y)^2=2x^2+2y^2+2xy-2x-2y+1
$$
Now, substitute $x=\hat x+a$ and $y=\hat y+b$. We get
$$
2\hat x^2+2\hat y^2+2\hat x\hat y+(4a+2b-2)\hat x+(2a+4b-2)\hat y+(2(a^2+ab+b^2-a-b)+1)
$$
We can eliminate the order 1 terms by letting $4a+2b-2=2a+4b-2=0$, which gives $a=b=\frac13$. With this substitution, we have
$$
2(\hat x^2+\hat x\hat y+\hat y^2)+\frac13
$$
We can express the bracketed term as a sum of squares, giving us
$$
3(\hat x+\hat y)^2+(\hat x-\hat y)^2 + \frac13
$$
and we can see that the smallest value this can take is $\frac13$. Indeed, it takes this value when $\hat x=\hat y=0$ - that is, when $x=y=\frac13$.
