Find a possible f(x) Do you think that it is possible to find a $f$ such that given a floating point constant $c \gt 0$ and an integer constant $n \gt 0$, then $\forall x_i \gt 1, i=1, 2,...,n$  :
$$f\left({1\over x_1}+{1\over x_2}+...+{1\over x_n}\right) = {1\over x_1 + c}+{1\over x_2 + c}+...+{1\over x_n + c}$$
 A: No.  We take the case $n=2$.  Note that $\frac 16+\frac 12=\frac 13 + \frac 13$, so you are asking that $$f(\frac 23)=\frac 1{2+c}+\frac 1{6+c}=\frac 2{3+c}, \text{ but}\\ \frac 1{2+c}+\frac 1{6+c}=\frac {8+2c}{12+8c+c^2}\\\frac 2{3+c}=\frac {8+2c}{12+7c+c^2}
$$ so they can only be equal if $c=0$  We can extend this to higher $n$ by adding the same set of fractions to each expression.
A: Update: Generalized for $n\geq3, c>0$, not just the case $n=3,c=1$.
For $n = 3, c>0, x_1 = x_2 = x_3 = 4c$:
\begin{gather}
\frac1{4c} + \frac1{4c} + \frac1{4c} = \frac3{4c}, \\
\frac1{4c+c} + \frac1{4c+c} + \frac1{4c+c} = \frac3{5c}.
\end{gather}
For $n = 3, c>0, x_1 = x_2 = 8c, x_3 = 2c$:
\begin{gather}
 \frac1{8c} + \frac1{8c} + \frac1{2c}   = \frac3{4c}, \\
\frac1{8c+c} + \frac1{8c+c} + \frac1{2c+c}  = \frac5{9c}.
\end{gather}
So for any possible function $f$,
$$
f\left( \frac1{4c} + \frac1{4c} + \frac1{4c} \right)
 = f\left( \frac3{4c} \right)
 = f\left( \frac1{8c} + \frac1{8c} + \frac1{2c} \right)
$$
but
$$
\frac1{4c+c} + \frac1{4c+c} + \frac1{4c+c}
 \neq \frac1{8c+c} + \frac1{8c+c} + \frac1{2c+c}.
$$
Hence for $n=3$ it is impossible that 
$f\left({1\over x_1}+{1\over x_2}+...+{1\over x_n}\right)$
can be equal to ${1\over x_1 + c}+{1\over x_2 + c}+...+{1\over x_n + c}$
in all cases; in at least one of these two cases the function
is not equal to the latter expression.
To extend this to $n>3$, 
let $x_4 = \cdots = x_n = 2$, and modify the sums above
by adding $\frac12 + \cdots + \frac12$ to the sums of $\frac{1}{x_i}$
and $\frac1{2+c} + \cdots + \frac1{2+c}$ to the sums of $\frac{1}{x_i + c}$;
we again have two equal sums of $\frac{1}{x_i}$ 
but unequal sums of $\frac{1}{x_i + c}$.
