<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] 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 
$$
3(x^2+y^2+z^2)\geq1
$$
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