# Prove that in an ordered field $(1+x)^n \ge 1 + nx + \frac{n(n-1)}{2}x^2$ for $x \ge 0$

In an ordered field show that $x \geq 0 \implies (1+x)^{n} \geq 1+nx+ \frac{1}{2}n(n-1)x^2$ for every positive integer $n$.

I know that $(1+x)^{n} \geq 1+nx$ (Bernoulli's inequality). To get the stronger inequality you can probably use induction also. But is there an easier way? The extra term on the right seems to be an "error" term and looks like a Taylor series expansion.

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it should be $x^2$. – Tom K Jun 20 '11 at 21:19
Yes, I would think that the $x^n$ you have should be $x^2$ instead. If it is then I guess you can just expand $(1 + x)^n$ using the Binomial theorem and... – Adrián Barquero Jun 20 '11 at 21:22
@Adrián: Come on... the binomial theorem is much harder than Bernoulli and this one :) – t.b. Jun 20 '11 at 21:24
But what's wrong with induction? It's just one line – t.b. Jun 20 '11 at 21:25
@Theo I guess the Binomial theorem is what seemed the most natural way to see where the inequality comes from in this case. And besides, since induction was ruled out is the only thing that came to mind ;) – Adrián Barquero Jun 20 '11 at 21:31

The following is an induction proof, and the OP wanted specifically to avoid induction, so it cannot be considered an answer. Actually, form a "logic" point of view, induction is unavoidable here.

Of course one can use the "Binomial Theorem." However, the context here is different, we are working in an ordered field. It turns out, if one does the details, that the standard induction proof of the Binomial Theorem in the familiar setting of the reals, or the complex numbers, can also be carried out in the general ordered field setting. But in principle one should then do it, and not simply claim that it can be done. So we need some information about binomial coefficients. Not hard.

So now we proceed to an induction proof of the inequality, in great detail. The result obviously holds when $n=1$. We want to show that if the result holds at $n=k$, then it holds at $n=k+1$. Thus we know that $$(1+x)^k \ge 1+kx + \frac{(k)(k-1)}{2}x^2 \qquad \text{(Inequality 1)}.$$

We wish to show that

$$(1+x)^{k+1} \ge 1+(k+1)x + \frac{(k+1)(k)}{2}x^2.$$

Multiply both sides of Inequality $1$ by the positive object $1+x$. Recall that in any ordered field, if $a \ge b$, and $c$ is positive, then $ac \ge bc$.

We find that $$(1+x)^{k+1} \ge \left(1+kx + \frac{(k)(k-1)}{2}x^2\right)(1+x)\qquad \text{(Inequality 2).}$$

Multiply out the right-hand side, gathering like powers, and doing some minor simplifications. We obtain $$1+(k+1)x + \left(\frac{(k)(k-1)}{2} +k\right)x^2 + \frac{(k)(k-1)}{2}x^3.$$

Finally, note that easy calculation shows that $(k)(k-1)/2 +k=(k+1)(k)/2$. Also, the term $((k)(k-1)/2)x^3$ is $\ge 0$, so we conclude from Inequality $2$ that

$$(1+x)^{k+1} \ge 1+(k+1)x + \frac{(k+1)(k)}{2}x^2,$$ which is what we wanted to show.

I think that the induction argument essentially writes itself, and is in that sense simple.

A confession: Despite the level of detail, there is a gap in the proof, actually several gaps of the same type.

In the solution, I casually simplified $kx+x$ as $(k+1)x$. In principle, one should justify that. When we write $kx$, we are not (quite) doing multiplication in the field. In principle, $kx$ simply means $x$ added to itself $k$ times. But then the result $kx+x=(k+1)x$ follows immediately. A similar comment applies anywhere in the computation where ordinary integers get mixed in with field elements, for example when one casually replaces $(kx)x$ by $kx^2$.

The above comments exemplify excessive fussiness. They should be noted and then one should go back to business as usual.

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Let $f(x)=(1 + x)^n$ and $g(x)=1 + nx + \frac{1}{2}n(n - 1)x^2$. Since $f(0)=g(0)=1$, it suffices to show that $f'(x) \geq g'(x)$, for any $x \geq 0$, in order to get $f(x) \geq g(x)$. Indeed, $f'(x) = n(1 + x)^{n - 1}$ and $g'(x)=n + n(n - 1)x$, and so it remains to show that $(1+x)^{n-1} \geq 1 + (n-1)x$, which you are given.

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I suppose we are not allowed to use this approach here. – Shai Covo Jun 20 '11 at 22:41
Nevertheless, it is interesting in its own right. – Shai Covo Jun 20 '11 at 22:45
I liked the observation that the stronger-than-Bernoulli inequality can be reduced to the Bernoulli inequality by differentiation. But since the question is about a general ordered field, is this actually a valid proof? Is that what you meant by your first comment? – Srivatsan Jul 30 '11 at 7:24
@Srivatsan: Actually, I didn't take much account of the "ordered field" issue. – Shai Covo Jul 31 '11 at 3:32
Nevertheless, this is a nice proof. +1 – Srivatsan Jul 31 '11 at 3:33

Since you only ask for $x \geq 0$, then just note that

$$(1 + x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k = 1 + nx + \frac{n(n-1)}{2} x^2 + \cdots$$

And I guess you can figure out the rest.

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@Theo Yeah I guess, I'm just going from memory since this is the way in which this inequality was derived when I was taking calculus II a few years ago and I liked it better than just having the inequality shown to me and proved by induction. Besides for me this is the best way to remember it since I just have to expand the first few terms of $(1+x)^n$ to get it. But I understand your comment though. – Adrián Barquero Jun 20 '11 at 21:38
Barquero: True, the Binomial Theorem could be proved in this setting. But the induction for that is surely more painful than the induction for the inequality. After doing that induction, one can remark that one could in fact prove a full Binomial Theorem, so as to round out the theoretical framework. – André Nicolas Jun 20 '11 at 21:40
@user6312 Yes sure, but I'm assuming that the OP has heard of the Binomial theorem and has probably seen it since he's asking about ordered fields. And I know that induction is easy, but he asked about other ways not involving induction. – Adrián Barquero Jun 20 '11 at 21:44