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. 
 A: 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.
A: 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.
A: 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.
