Find $x,y,z>0$ such that $x+y+z=1$ and $x^2+y^2+z^2$ is minimal How can I find $3$ positive numbers that have a sum of $1$ and the sum of their squares is minimum?
So far I have:
$$x+y+z=1 \qquad \implies \qquad z=1-(x+y)$$
So,
$$f(x,y)=xyz=xy(1-x-y)$$
But I'm stuck from here.  Hints?

By C-S $$3(x^2+y^2+z^2)=(1+1+1)(x^2+y^2+z^2)\geq(x+y+z)^2=1.$$
Thus, $x^2+y^2+z^2\geq\frac{1}{3}$.
The equality occurs for $x=y=z=\frac{1}{3}$, which is also the answer.
