Recently, when I self-learned Discrete Mathematics and Its Applications 8th by Kenneth Rosen, I did only the even-numbered exercises which the author offers the detailed description than the odd ones because the odd ones and the corresponding even ones are very similar.
I was stuck at exercise 5-Supplementary-36 which is related with induction. The problem is
Use mathematical induction to prove that if $x_1, x_2, \ldots , x_n$ are positive real numbers with $n\ge 2$, then $$ (x_1+\frac{1}{x_1})(x_2+\frac{1}{x_2})\cdots(x_n+\frac{1}{x_n})\ge (x_1+\frac{1}{x_2})(x_2+\frac{1}{x_3})\cdots(x_n+\frac{1}{x_{n-1}})(x_1+\frac{1}{x_n}) $$
The answer says
It will be convenient to clear fractions by multiplying both sides by the product of all the $x_s$’s ; this makes the desired inequality $$ (x_1^2 + 1)(x_2^2 + 1)\cdots(x_n^2 + 1) \ge (x_1 x_2 + 1)(x_2 x_3 + 1) ···(x_{n−1}x_n + 1)(x_n x_1 + 1). $$ ...
Because of the cyclic form of this inequality, we can without loss of generality assume that $x_{n+1}$ is the largest (or tied for the largest) of all the given numbers
Q:
However, when I swap between arbitrary $x_i,x_j$ where $1\le i,j\le n$, then $(x_1 x_2 + 1)(x_2 x_3 + 1) ···(x_{n−1}x_n + 1)(x_n x_1 + 1)$ will change value. Then how does the above the WLOG(without loss of generality) assumption work? If the assumption has some problems, how to prove the above theorem?
Let $f(x_1, x_2, \ldots, x_n)$ denote the difference between left-hand side and right-hand side of the desired inequality. Then $$ f(x_1, x_2, \ldots, x_n) = f(x_2, \ldots, x_n, x_1) \, , $$ i.e. the function is invariant under a cylic permutation of its arguments.
With a finite number of such cyclic permutations the largest (or smallest) of the arguments can be moved to the first (or last) position, as desired.
It is not claimed that $f$ is invariant under an arbitrary permuation of the arguments (which is a much stronger property).