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?

$$f(x,y)=x^2+y^2+(1-x-y)^2$$
$$\frac{\partial f}{\partial x}=2x+2(1-x-y)(-1)=0\tag 1$$
$$\frac{\partial f}{\partial y}=2y+2(1-x-y)(-1)=0\tag 2$$
by subtracting 
$$x=y$$ 
substitute in 1 to get
$$x=\frac{1}{3}$$ 
so
$$y=\frac{1}{3}$$
and then $$z=1-x-y=\frac{1}{3}$$ 
to check the critical point, we can reduce the original function, which has three variables, to one varible as follow
$$f(x)=x^2+x^2+(1-x-x)^2=2x^2+(1-2x)^2$$ 
it is very easy to find the critical point if there is one variable
