<s>[INST] 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?
 [/INST] 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 ${1\over3}\le x^2+y^2+z^2$.
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