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?

$x \mapsto x^2$ is a convex function. By Jensen's inequality,
$$x^2+y^2+z^2 = 3\left(\frac{x^2+y^2+z^2}{3}\right) \ge 3\left(\frac{x+y+z}{3}\right)^2 = 3\left(\frac{1}{3}\right)^2 = \frac13$$
Since the equality $x^2+y^2+z^2 = \frac13$ is achieved at $x = y = z = \frac13$, this is the solution you seek. 
If you want a more elementary approach, you can use the fact
$$\begin{align}
x^2 + y^2 + z^2 
=& \left(x-\frac13+\frac13\right)^2 + \left(y-\frac13+\frac13\right)^2 + \left(z -\frac13 + \frac13\right)^2\\
=& \left(x-\frac13\right)^2 + \left(y-\frac13\right)^2 + \left(z -\frac13\right)^2\\
&+ \frac23\left[\left(x-\frac13\right) + \left(y-\frac13\right) + \left(z -\frac13\right)\right] + 3\left(\frac13\right)^2\\
=& \left(x-\frac13\right)^2 + \left(y-\frac13\right)^2 + \left(z -\frac13\right)^2 + \frac13
\end{align}
$$
to arrive at same conclusion.
