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Note: I don't want to get the full solution, but only a hint.

I have to show that for $x \in [0, \infty)$, the sequence $\left(1+\frac xk\right)^k$ is monotonically increasing. We were already given the hint that we could use the inequality of arithmetic and geometric means, but I don't see how to apply it yet.

I tried to do it by induction. While the case $k = 1$ works easily, I don't know how to start further from here. I already tried to use the definition of the binomial theorem, but it didn't lead me anywhere.

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Hint:

Apply AM - GM inequality with $m > k$ to:

$$\left(1+\frac{x}{k}\right)^{k/m} =\left(\frac{k+x}{k}\right)^{k/m} = \left[\left(\frac{k+x}{k}\right)^k \right]^{1/m} \\= \left[\underbrace{\frac{k+x}{k} \ldots \frac{k+x}{k}}_k \underbrace{1 \ldots 1}_{m-k}\right]^{1/m} \\ \leqslant \frac{1}{m}\left(k \frac{k+x}{k } + (m-k)(1) \right)$$

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  • $\begingroup$ Still by induction? Or "directly"? $\endgroup$ – Julian Dec 10 '16 at 11:22
  • $\begingroup$ Directly. Complete the arithmetic mean on the right side and see how it simplifies. $\endgroup$ – RRL Dec 10 '16 at 11:24
  • $\begingroup$ After that you do some exponentiation to both sides and you should be done $\endgroup$ – RRL Dec 10 '16 at 11:27
  • $\begingroup$ But how do you write an arbitrary $k$ as $n \over m$ with $m > n$, if I may ask? It seems like you used that here. $\endgroup$ – Julian Dec 10 '16 at 11:29
  • $\begingroup$ My $n$ is your $k$. I'm showing for any $m$ greater than $n$ that $(1 + x/n)^n \leqslant (1+x/m)^m$. The intermediate step is $(1+x/n)^{n/m} \leqslant (1 + x/m)$. $\endgroup$ – RRL Dec 10 '16 at 11:33
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The AM-GM tells us that for non negative $\;a_n\;$ :

$$\frac{a_1+\ldots+a_n}n\ge\sqrt[n]{a_1\cdot\ldots\cdot a_n}$$

Let us apply this with

$$a_1=1\,,\,\,a_2=a_3=\ldots=a_{n+1}=1+\frac xn\;,\;\;\text{and then get :}$$

$$\frac{a_1+\ldots+a_{n+1}}{n+1}\ge\sqrt[n+1]{a_1\cdot\ldots\cdot a_{n+1}}\iff$$ $$\frac{1+\left(1+\frac xn\right)+\ldots+\left(1+\frac xn\right)}{n+1}\ge\sqrt[n+1]{1\cdot\left(1+\frac xn\right)\cdot\ldots\cdot\left(1+\frac xn\right)}\stackrel{\text{raise to power}\;n+1 }\iff$$

$$\left(\frac{1+n\left(1+\frac xn\right)}{n+1}\right)^{n+1}\ge\left(1+\frac xn\right)^n\iff\left(1+\frac x{n+1}\right)^{n+1}\ge\left(1+\frac xn\right)^n$$

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    $\begingroup$ Beautiful approach. Thank you. By far the clearest exposition of this fact that I've seen. $\endgroup$ – user370967 Nov 29 '18 at 21:11
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For $x=0$ the function is constant, so we can assume $x>0$.

Consider $$ f(t)=\frac{\log(1+xt)}{t} $$ for $t>0$. Then $$ f'(t)=\frac{t\frac{x}{1+xt}-\log(1+xt)}{t^2}= \frac{xt-(1+xt)\log(1+xt)}{t^2(1+xt)} $$ Now our task is proving that, for $u>0$, $$ g(u)=u-(1+u)\log(1+u)<0 $$ We indeed have $g(0)=0$ and $$ g'(u)=1-\log(1+u)-1=-\log(1+u) $$

Put together the pieces and conclude.

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Expand it using the binomial theorem and try to look at the sum term by term. Create a sequence of each term (other than 1 obviously )and try to get that it increases for increasing $k $.

The sum will be $1+^kC_1(\frac xk) + ^kC_2 (\frac xk)^2 + ... ^kC_k (\frac xk)^k $.This could be written as $1+\sum_{r=1}^k {^kC_r(\frac xk)^r}$ Now consider the sequence $^kC_r(\frac 1k)^r$ and try to prove that this sequence is increasing for increasing $k$ for any $r >=1$. This will prove that each term of the sum increases thus implying the sum increases .

One way to prove that the sequence $ S = ^kC_r(\frac 1k)^r$ increase is to calculate the ratio $S (k+1)/S (k) $ and prove that is greater than or equal to 1 for all $k $.

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