<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] By Cauchy-Schwarz, for every $x,y,z\in\Bbb R$ we have
$$x+y+z=\langle(1,1,1),(x,y,z)\rangle \leq \left\|(1,1,1)\right\|_2\cdot \|(x,y,z)\|_2=\sqrt{3}\cdot \sqrt{x^2+y^2+z^2}$$
Now, if $x+y+z=1$, then squaring both sides and dividing by $3$ gives
$$ \frac{1}{3}\leq x^2+y^2+z^2 $$
This lower bound is then attained for $|x|=|y|=|z|=1/3$. 
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