# Sum of three numbers and the sum of their squares: $x^2+y^2+z^2 \geq \frac{1}{3}$ for $x+y+z=1$

We have $x, y, z \in \mathbb R$ such that $x+y+z=1$. Prove that $x^2+y^2+z^2 \geq \frac{1}{3}$. I am able to do this using the relationship between the power and arithmetic means. Is there a way to not use this relationship?

Hint: $(x-\frac13)^2 + (y-\frac13)^2 + (z-\frac13)^2 \geq 0$.
We have $$x+y+z=1\\ (x+y+z)^2=1\\ x^2+y^2+z^2+2(xy+xz+yz)=1$$ By the rearrangement inequality, $x^2+y^2+z^2\geq xy+xz+yz$. Inserting that gives you $$3(x^2+y^2+z^2)\geq1$$
For any real numbers $x,y,z$, we have
\begin{align} 0\le(x-y)^2+(y-z)^2+(z-x)^2&\implies2(xy+yz+zx)\le2(x^2+y^2+z^2)\\ &\implies(x+y+z)^2\le3(x^2+y^2+z^2) \end{align}
So if $x+y+z=1$, then ${1\over3}\le x^2+y^2+z^2$.