<s>[INST] 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?
 [/INST] Lagrange multipliers? So $f(x,y,z,\lambda) = x^2 + y^2 + z^2 - \lambda(x+y+z-1)$.
So critical points obey: $2x - \lambda = 0, 2y - \lambda = 0, 2z - \lambda = 0, x+y+z-1 = 0$. So $x = y = z = \frac{1}{3}$, which gives sum of squares $\frac{1}{3}$. 
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