Let the three positive real numbers be x,y and z.

then, x+y+z=1

let sum of squares be:

\(\displaystyle{f{{\left({x},{y},{z}\right)}}}={x}^{{2}}+{y}^{{2}}+{z}^{{2}}\)

put z in above function

\(\displaystyle{g{{\left({x},{y}\right)}}}={x}^{{2}}+{y}^{{2}}+{\left({1}-{\left({x}+{y}\right)}\right)}^{{2}}\)

\(\displaystyle={x}^{+}{y}^{{2}}+{1}+{\left({x}+{y}\right)}^{{2}}-{2}\cdot{1}\cdot{\left({x}+{y}\right)}\)

\(\displaystyle={x}^{{2}}+{y}^{{2}}+{1}+{2}^{{2}}+{y}^{{2}}+{2}{x}{y}-{2}{x}-{2}{y}\)

\(\displaystyle={2}{x}^{{2}}+{2}{y}^{{2}}+{2}{x}{y}-{2}{x}-{2}{y}+{1}\)

Partially differenting the function g(x,y) and finding the critical points:

\(\displaystyle\frac{{\partial{g}}}{{\partial{x}}}={4}{x}+{2}{y}-{2}={0}\)

\(\displaystyle\frac{{\partial{g}}}{{\partial{y}}}={4}{y}+{2}{x}-{2}={0}\)

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:

\(\displaystyle{g}_{{\times}}={4},\)

\(\displaystyle{g}_{{{y}{y}}}={4},\)

\(\displaystyle{g}_{{{x}{y}}}{4}{x}+{2}={4}{\left(\frac{{1}}{{3}}\right)}+{2}=\frac{{10}}{{3}}\)

\(\displaystyle{D}={g}_{{\times}}\cdot{g}_{{{y}{y}}}-{{g}_{{{x}{y}}}^{{2}}}\)

\(\displaystyle={4}\cdot{4}-{\left(\frac{{10}}{{3}}\right)}^{{2}}\)

\(\displaystyle={16}-\frac{{100}}{{9}}\)

\(\displaystyle=\frac{{44}}{{9}}{>}{0}\)

Since \(\displaystyle{g}_{{\times}}{>}{0}{\quad\text{and}\quad}{D}{>}{0},{p}{\left(\frac{{1}}{{3}},\frac{{1}}{{3}},\frac{{1}}{{3}}\right)}\) is a local minimal

Therefore, the sum of squarers will be minimom if \(\displaystyle{x}={y}={z}=\frac{{1}}{{3}}\) Hence, \(\displaystyle{x}=\frac{{1}}{{3}},{y}=\frac{{1}}{{3}},{z}=\frac{{1}}{{3}}.\)

then, x+y+z=1

let sum of squares be:

\(\displaystyle{f{{\left({x},{y},{z}\right)}}}={x}^{{2}}+{y}^{{2}}+{z}^{{2}}\)

put z in above function

\(\displaystyle{g{{\left({x},{y}\right)}}}={x}^{{2}}+{y}^{{2}}+{\left({1}-{\left({x}+{y}\right)}\right)}^{{2}}\)

\(\displaystyle={x}^{+}{y}^{{2}}+{1}+{\left({x}+{y}\right)}^{{2}}-{2}\cdot{1}\cdot{\left({x}+{y}\right)}\)

\(\displaystyle={x}^{{2}}+{y}^{{2}}+{1}+{2}^{{2}}+{y}^{{2}}+{2}{x}{y}-{2}{x}-{2}{y}\)

\(\displaystyle={2}{x}^{{2}}+{2}{y}^{{2}}+{2}{x}{y}-{2}{x}-{2}{y}+{1}\)

Partially differenting the function g(x,y) and finding the critical points:

\(\displaystyle\frac{{\partial{g}}}{{\partial{x}}}={4}{x}+{2}{y}-{2}={0}\)

\(\displaystyle\frac{{\partial{g}}}{{\partial{y}}}={4}{y}+{2}{x}-{2}={0}\)

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:

\(\displaystyle{g}_{{\times}}={4},\)

\(\displaystyle{g}_{{{y}{y}}}={4},\)

\(\displaystyle{g}_{{{x}{y}}}{4}{x}+{2}={4}{\left(\frac{{1}}{{3}}\right)}+{2}=\frac{{10}}{{3}}\)

\(\displaystyle{D}={g}_{{\times}}\cdot{g}_{{{y}{y}}}-{{g}_{{{x}{y}}}^{{2}}}\)

\(\displaystyle={4}\cdot{4}-{\left(\frac{{10}}{{3}}\right)}^{{2}}\)

\(\displaystyle={16}-\frac{{100}}{{9}}\)

\(\displaystyle=\frac{{44}}{{9}}{>}{0}\)

Since \(\displaystyle{g}_{{\times}}{>}{0}{\quad\text{and}\quad}{D}{>}{0},{p}{\left(\frac{{1}}{{3}},\frac{{1}}{{3}},\frac{{1}}{{3}}\right)}\) is a local minimal

Therefore, the sum of squarers will be minimom if \(\displaystyle{x}={y}={z}=\frac{{1}}{{3}}\) Hence, \(\displaystyle{x}=\frac{{1}}{{3}},{y}=\frac{{1}}{{3}},{z}=\frac{{1}}{{3}}.\)