Using Mathematica to confirm Bernoulli's inequality
Since Reduce
doesn't seem to like the inequality, I tried FullSimplify
with Assumptions
instead. This works in three steps:
differenceByTerm =
SeriesCoefficient[(1 + x)^n - (1 + n x), {x, 0, m}]
$$ \cases{ 0 & m=0 \\ \binom{n}{m} & m>1 \\ 0 & \text{True} \\ }$$
FullSimplify[
differenceByTerm >= 0,
Assumptions -> n > 1 && {m, n} \[Element] Integers && n >= m > 1]
(* ==> True *)
FullSimplify[
differenceByTerm >= 0,
Assumptions -> n > 1 && {m, n} \[Element] Integers && m > n]
(* ==> True *)
So I did the comparison of the two sides term by term in an expansion in powers of x
. This is doable because SeriesCoefficient
, unlike Coefficient
, allows symbolic powers. So the result differenceByTerm
is a function of the degree n
of the polynomial, and the power m
in the expansion. It's a case distinction, where the True
entry refers to the special case m==1
.
Finally, I have to test whether this difference is larger or equal to zero. This is a little easier than the strict inequality in the original question.
But FullSimplify
only manages to decide this if I treat the cases $m>n$ and $m\le n$ separately. In both cases, the result is True
, so the statement is proved.
Here's an inductive proof:
Defining the function
f[n_] := (1 + x)^n - (1 + n x)
we have to prove that f[n] > 0
for n > 1
and x > -1
.
Now we observe that for f[n] we have the identity
Simplify[f[n + 1] == (1 + x) f[n] + n x^2]
(*
True
*)
It is obvious "by eye" that f[n] > 0
because there are only positive quantities involved on the right hand side.
It is easily proved formally that the expression for a similar function g[n+1]
is positive provided g[n]
is:
Simplify[(1 + x) g[n] + n x^2 > 0, {x > -1, n > 1, g[n] > 0}]
(*
Out[2]= True
*)
Notice that the proof holds for real n > 1
, so that it is more general than requested in the OP.
Remark 1
RSolving the recursion eq1 = g[n + 1] == (1 + x) g[n] + n x^2 with g[1] = 0 brings us back to f[n] and is therefore of no use.
Remark 2
RSolving the modified recursion
eq2 = h[n + 1] == (1 + x) h[n];
with
h[2] == x^2
gives
h[n] = x^2 (1 + x)^(-2 + n) for n >= 2
which Mathematica recognizes to be positive:
Simplify[x^2 (1 + x)^(-2 + n) > 0, {x > 0, n > 0}]
(*
Out[1]= True
*)
If we could prove that h[n] is smaller than f[n] we are done because h > 0. But I haven't found that proof.