Find a possible f(x)

Question

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 $\forall x_i \gt 1, i=1, 2,...,n$ :

$$f\left({1\over x_1}+{1\over x_2}+...+{1\over x_n}\right) = {1\over x_1 + c}+{1\over x_2 + c}+...+{1\over x_n + c}$$

Answer 1

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.

Answer 2

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 $\frac{1}{x_i + c}$.