Note that the inequality

          b f(a) + a f(b)
 f(a)  <  ---------------  <  f(b)  if  f(a) < f(b), a>0, b>0
               a + b

is a special case of

  X    <  (1-t) X  +  t Y  <  Y,    if    X  <  Y,   t>0, t<1

i.e. the above middle term is on the "line" between X and Y.

Also, the proof as written fails if a or b is 0, but it is
easily remedied since  f(0) = f(1)  (put x = 0, y = 1), so
that we may assume a,b nonzero (or set N = {1,2,3,...} ).

That N is discrete and not dense is a nice way to recast many
induction or infinite descent proofs, e.g. irrationality proofs:

From: Bill Dubuque <w...@zurich.ai.mit.edu>
To: historia-matemat...@chasque.apc.org
Subject: Re: [HM] sqrt(2) and reductio
Date: Sat, 16 Jun 2001 19:38:40 -0400

Below is one of my favorite irrationality proofs. It vividly
highlights the importance of the notion of algebraic integer,
and works by contrasting the DISCRETENESS of a ring of integers
versus the DENSENESS of a ring of fractions. I have presented
the proof in a very simple form - high-school algebra suffices.

Has anyone seen this proof before?  If so, what is its history?

THEOREM.  Let Z denote the integers. If an integer b in Z
has a rational sqrt: sqrt(b) = c/d, then c/d is an integer.

PROOF. Suppose, as hypothesized, w = sqrt(b) = c/d is rational.
d is a common denominator for all elts in the ring  R = Z[w]
since if  r = m + nw in R, m,n in Z,  dr = dm + n(dw) is in Z
via  dw = c in Z. So R is a subset of Z/d, i.e. every  r in R
has form n/d for an integer n. Suppose r = n/d is nonintegral.
We will show this leads to a contradiction and, hence, that
every r in R = Z[w] must be an integer, including w = sqrt(b).

W.l.o.g we may assume 0 < r < 1 by taking the fractional part
of |r|, since this too lies in R and is nonintegral iff r is.
Since r has form  n/d  and  0 < r < 1,  r is a member of the
finite set { 1/d, 2/d,...,(d-1)/d }.  If r is the smallest
elt of R in this set then  r^2  is an even smaller such elt
since  1 > r > r^2 > 0, contradicting the minimality of r. QED

This proof works for any algebraic *integer*, i.e. a root of
p(w) = w^n + a w^(n-1) + b w^(n-2) + ... + k,  a,b,..,k in Z,
using  d^(n-1)  versus  d  as a common denominator for Z[w].
It proves any rational algebraic integer is an integer (in Z).
The proof above is simply the special case  p(w) = w^2 - b.

Note the subtlety here. The proof depends crucially on the fact
that w is an algebraic *integer*, which implies that the ring
 Z[w] = Z<1,w,w^2,..,w^(n-1)>  is a finite dim Z-module, i.e.
w^n, w^(n+1),.. are Z-linear combin's of  1, w, w^2,.., w^(n-1)
This needn't be true if w is a root of a poly p(w) whose leading
coef c is not unity since then  w^n = -1/c (a w^(n-1) + ... + k)
so that c may intrude as a nontrivial denominator when using this
equation to rewrite  w^n, w^(n+1),... after multiplying two elts.

Essentially the proof shows that if R > Z is an ordered ring
extension which adjoins a new finite element then R is dense,
having infinite descending chains  1 > r > r^2 > r^3 > .. > 0
in a nbhd of 0 (hence everywhere by an additive shift), where
r > 0 is the fractional part of any newly adjoined finite elt.
But  Z/d  is discrete, not dense, hence the contradiction.

Note that I say "finite" above because it is possible to adjoin
infinite elts alone, e.g. consider Z[x] with x infinite, i.e.
poly's in a nbhd of positive infinity = +oo, ordered by declaring
 p(x) > q(x) iff this holds true eventually for all large x > 0,
i.e. in a nbhd of +oo.  This is same as declaring p(x) positive
or negative as is its leading coef (the leading term eventually
dwarfs others as x-> +oo), and  p > q  <=>  p - q > 0, as usual.
This ordered ring extension of Z adjoins no new finite elements.
However, if we additionally adjoin the infinitesimal  1/x  then
we have  1 > 1/x > 1/x^2 > ... > 0, and adding c shifts this
infinitely descending chain to the nbhd of any elt c in Z[x,1/x].

Source: http://mathforum.org/epigone/historia_matematica/gosneyah/E15BPeK-0003Mh...@nestle.ai.mit.edu