28*. Variables \(x\) and \(y\) are positive, \(x+y=6\). Find the minimum value of the sum \(\frac{1}{x}+\frac{1}{y}\).
28. Solution. First method. Since \(y=6-x\), the given sum is:

\[
\frac{x+y}{x y}=\frac{6}{x(6-x)}=\frac{6}{9-9+6 x-x^{2}}=\frac{6}{9-(3-x)^{2}}
\]

The sum will be the smallest when \((3-x)^{2}=0\), i.e., when \(x=3\), so the smallest value of the sum is \(\frac{2}{3}\).

Second method. Let \(x=3+a\), then \(y=6-3-a=3-a\). Performing the corresponding substitution, we get:

\[
\frac{1}{3+a}+\frac{1}{3-a}=\ldots=\frac{6}{9-a^{2}}
\]

Obviously, the sum will be the smallest when \(a=0\) and so on.

Third method. One can use the identity:

\[
(x+y)^{2}-(x-y)^{2}=4 x y
\]