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The question reads

$$U_n = (1+x)^n - 1 - nx$$ Show that $U_2 \geq 0$ Hence or otherwise show that $(1+x)^n \geq 1 + nx$ for all $x \gt -1$.

Obviously the $U_2 \geq 0$ is very easy, I can do that without any trouble but I cannot see how it links to the 2nd part of the question.

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I think induction on $n$ would do the trick here, which means you need to show it holds for $n=2$; note that $U_2\geq 0$ is equivalent to showing $(1+x)^2\geq 1+2x$, which holds for all $x$, whether or not they are greater than $-1$.

Now, assume that $U_n\geq 0$; that is, that $(1+x)^n\geq 1+nx$, and that $x\gt -1$. We want to prove that $U_{n+1}\geq 0$ (i.e., that $(1+x)^{n+1}\geq 1+(n+1)x$).

Take $(1+x)^n\geq 1+nx$. Since $x\gt -1$, then $1+x\gt 0$; multiplying both sides by $1+x$ we get: $$\begin{align*} (1+x)^n &\geq 1+nx\\ (1+x)^n(1+x)&\geq (1+nx)(1+x)\\ (1+x)^{n+1}&\geq 1 + nx + x + nx^2\\ (1+x)^{n+1}&\geq 1+ (n+1)x + nx^2 \geq 1+(n+1)x \end{align*}$$ with the last inequality since $x^2\geq 0$, so $nx^2\geq 0$. This proves that if $(1+x)^n\geq 1+nx$ and $x\gt -1$, then $(1+x)^{n+1}\geq 1+(n+1)x$. Since $(1+x)^2\geq 1+2x$, then the result holds for all $n$ by induction.

(This can also be done by calculus, noting that $f(x)=(1+x)^n$ lies above its tangent at $x=0$ on the interval $(-1,\infty)$; the tangent at $0$ is precisely $y=1+nx$.)

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Thanks for the explanation, I was thinking of doing it through induction too but what gets me is the word "hence". I feel there the first part should lead into doing it. I know it does say "otherwise" and doing it by induction is perfectly acceptable but I still cannot see how to answer it in the way the question wants. Thanks for your help though – Mark Dunne Jun 9 '11 at 21:01
@Max: I don't understand the "hence or otherwise" either. – Arturo Magidin Jun 9 '11 at 21:02

HINT $\rm\ \ U_{n+1}\ =\ (x+1)\ U_n + n\ x^2\ $ which is $\:\ge 0\:$ by induction.

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Instead of proving this, I'll prove a modest generalisation. Let $x_1,x_2, \ldots x_n$ be numbers greater than $-1$, then $(1+x_1)(1+x_2)\ldots(1+x_n)\geq 1+x_1+x_2+\ldots x_n$

The inductive step is:

$(1+x_1)(1+x_2)\ldots(1+x_{n+1})\geq(1+x_1+x_2+\ldots+x_n)(1+x_{n+1})$ $=\sum_1^{n+1}x_i+x_{n+1}(\sum_1^n x_i)\geq \sum_1^{n+1}x_i$

Bernoulli's inequality follows when all the $x_i$'s are the same.

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Thanks for your help, but I'll say the same as I did to Arturo Magidin, I know there is no problem in doing this by induction but the question really hints at doing it as a follow on step from proving U2 >= 0. Thanks for the help though – Mark Dunne Jun 9 '11 at 21:08
Hi. The 'hence' part was proved by Arturo Magidin, I proved the 'otherwise' bit. – Dactyl Jun 9 '11 at 21:11

My favorite way of proving Bernoulli is to use Jensen inequality. First of all, the inequality is trivial if $1+nx\leq 0$. So suppose that $1+nx>0$. Following inequality can be proved using Jensen inequality and the fact that $\log$ function is concave: $$ \frac 1n\log(1+nx)+\frac{n-1}{n}\log 1\leq \log(\frac 1n(1+nx)+\frac{n-1}n)=\log(1+x), $$ which is the desired inequality. As a matter of fact it does not matter if $n$ is integer here. It suffices that $n\geq 1$ and it is a real number.

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