Use the arithmetic mean-geometric mean inequality to establish the following results:

(a) If $nt > -1 $, then $(1-t)^{n} \geq 1 - nt$

(b) If $-x < n < m$, then $(1+x/n)^{n} \leq (1 + x/m)^{m}$


(a) I had no success at trying to prove it using the AM-GM inequality. Thus I've tried to apply induction.

The base case is true, since for $n = 1$ we get that $(1-t)^{1} = 1 - t = 1 - 1\cdot t$.

Now let us assume that it holds for some natural number $n\geq 1$ and we shall prove it holds for $n + 1$ too.

Indeed, one has that \begin{align*} (1-t)^{n+1} = (1-t)^{n}(1-t) \geq (1-nt)(1-t) = 1 - (n+1)t + nt^{2} \geq 1 - (n+1)t \end{align*} and we are done.

(b) Similarly, we shall prove it applying induction on $m$. For $m = 2$ it is true, because \begin{align*} n < 2 \Rightarrow n = 1 \Rightarrow \left(1 + \frac{x}{2}\right)^{2} = 1 + x + \frac{x^{2}}{4} \geq 1 + x \end{align*}

But then I get stuck.

Could someone please finish the induction proof and provide a solution based on AM-GM inequality?

Any contribution is appreciated.

  • $\begingroup$ You can not use induction here or you need to say that $m$ and $n$ are natural numbers. $\endgroup$ Commented Jul 8, 2020 at 1:27

1 Answer 1


We need also $n>0$, otherwise your inequality is wrong.

For $n>0$ since $$\frac{n}{m}+\frac{m-n}{m}=1,$$ By AM-GM we obtain $$1+\frac{x}{m}=\frac{n}{m}\left(1+\frac{x}{n}\right)+\frac{m-n}{m}\geq\left(1+\frac{x}{n}\right)^{\frac{n}{m}}\cdot1^{\frac{m-n}{m}}=\left(1+\frac{x}{n}\right)^{\frac{n}{m}},$$ which gives your inequality.


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