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.