<s>[INST] 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.
 [/INST] Ok, you want an intuitive proof, not a better proof. That's fine.
First note that for $x=y=z$, we have $x^2+y^2+z^2 = 3\left(\frac13\right)^2= \frac13$. Next let us show that $x^2+y+2+z^2$ is not minimal if $x\ne y$ or $y\ne z$. Indeed if $a\ne b$ then
$$
a^2 + b^2 - \left(\left(\frac{a+b}2\right)^2+\left(\frac{a+b}2\right)^2\right) = \frac12\left(
a^2 + b^2 -2ab
\right)=\frac12(a-b)^2>0,
$$
so we can replace two of the numbers be their averages and lower the sum of squares.
Unfortunately we not yet quite done. We are done if we can show that there is a global minimum of $x^2+y^2+z^2$, given that $x+y+z=1$. Because then that global minimum must also be a local minimum, but for that we cannot have $x\ne y$ or $y\ne z$, so $x=y=z$ must be the global minimum, and its value is $\frac13$.
For that we can first argue that we can restrict ourselves to $x,y,z\ge0$ (e.g. if $x<0$ replace $(x,y,z)$ by $\frac1{-x+y+z}(-x,y,z)$, which yields a smaller value since $-x+y+z = 1 - 2x > 1$) and then appeal to the compactness of $\{(x,y,z)\colon \text{$x+y+z=1$, $x,y,z\ge0$}\}$.
I did not say that this would be pretty, but it is pretty intuitive to me :)
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