Multivariable optimization question. Find three positive real numbers whose sum is one and the sum of their squares is a minimum.

Let the three positive real numbers be x,y and z.
then, $x+y+z=1$
let sum of squares be:
${\displaystyle f(x,y,z)=x^2+y^2+z^2}$
put z in above function
${\displaystyle g(x,y)=x^2+y^2+{(1-(x+y))}^2}$
${\displaystyle =x^{+}{y}^2+1+{(x+y)}^2-2\cdot 1\cdot (x+y)}$
${\displaystyle =x^2+y^2+1+2^2+y^2+2xy-2x-2y}$
${\displaystyle =2x^2+2y^2+2xy-2x-2y+1}$
Partially differenting the function g(x,y) and finding the critical points:
${\displaystyle \frac{\partial g}{\partial x}=4x+2y-2=0}$
${\displaystyle \frac{\partial g}{\partial y}=4y+2x-2=0}$
From the above two equtioan, we get:
${\displaystyle x=\frac{1}{3},y=\frac{1}{3}}$
Putting the value of x and y in $x+y+z=1$, we get, ${\displaystyle z=\frac{1}{2}}$
Using the second derivative test:
${\displaystyle g_{\times }=4,}$
${\displaystyle g_{yy}=4,}$
${\displaystyle g_{xy}4x+2=4\left(\frac{1}{3}\right)+2=\frac{10}{3}}$
${\displaystyle D=g_{xx}\cdot g_{yy}-g_{xy}^2}$
${\displaystyle =4\cdot 4-{\left(\frac{10}{3}\right)}^2}$
${\displaystyle =16-\frac{100}{9}}$
${\displaystyle =\frac{44}{9}>0}$
Since ${\displaystyle g_{xx}>0\text{and}D>0,p(\frac{1}{3},\frac{1}{3},\frac{1}{3})}$ is a local minimal
Therefore, the sum of squarers will be minimom if ${\displaystyle x=y=z=\frac{1}{3}}$ Hence, ${\displaystyle x=\frac{1}{3},y=\frac{1}{3},z=\frac{1}{3}.}$