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I need to show that $e < \Big(1 + \frac 2 {2x+1}\Big)^{x+1}$ for all $x \ge 1$.

This happens if $(x+1)\ln\Big(1 + \frac 2 {2x+1}\Big) > 1$ so let's study the function $f(x) = (x+1)\ln\Big(1 + \frac 2 {2x+1}\Big)$

Edit

We have abandoned the previous method of only using a characterization of $e$ with inequalities since we thought it was impractical to do it. Let's return instead to analysis.

  1. $f(1) = 2 \ln(\frac 5 3) > 1$
  2. $f'(x) = \ln(1+ \frac 2 {2x+1}) - \frac {4(x+1)}{(4x^2+8x+3)} < 0$ on $[1,+\infty[$
  3. $\lim f(x) = 1$

However how do I show formally 2.?

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  • $\begingroup$ Reverting to the previous version: you made the exponent bigger in the same way but you made the inside smaller, and so these two inequalities alone shouldn't really be able to tell you which way the inequality will go. $\endgroup$ – Ian May 30 '19 at 13:17
  • $\begingroup$ Yes, somehow I was comparing $2x+1$ to $2x+2$ even though that makes no sense, it should be compared to $2x$. $\endgroup$ – Ian May 30 '19 at 13:17
  • $\begingroup$ @Ian so you think there is no hope? $\endgroup$ – Rodrigo May 30 '19 at 13:19
  • $\begingroup$ I don't think you can do it, since you are essentially asking to determine whether the geometric mean of the upper bound and lower bound is an upper/lower bound in the absence of any other information. $\endgroup$ – user10354138 May 30 '19 at 13:19
  • $\begingroup$ With no other tools whatsoever, this won't be enough; you've just picked a number between two bounds for $e$ and you ask which side of $e$ that number is on, you'll need more information to do that. $\endgroup$ – Ian May 30 '19 at 13:20
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Hint: $$ \Big(1 + \frac 2 {2x+1}\Big)^{x+1}= \Big(1 + \frac{1+\frac{1}{2x+1}}{x+1}\Big)^{x+1}>\Big(1 + \frac{1}{x+1}\Big)^{x+1}$$

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    $\begingroup$ The right-hand side above is less the $e$. So the estimate is not a full proof of the original inequality. $\endgroup$ – StarBug May 31 '19 at 1:30
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Consider the function $f(y):=(1+\frac{2}{y})^{y+1}$. Clearly, $\lim_{y\rightarrow\infty}f(y)=e^2$. Moreover, $f$ is monotone decreasing since \begin{align*} f'(y)<0 &\iff \left({1+\frac{2}{y}}\right)\log\left({1+\frac{2}{y}}\right) < \left({\frac{2}{y}}\right)+\frac{1}{2}\left({\frac{2}{y}}\right)^2 \\ & \iff (1+t)\log(1+t) < t+\frac{1}{2}t^2 \quad \text{for }t=\frac{2}{y}, \end{align*} where the last inequality can be verified for all $t> 0$ since equality holds for $t=0$ and the right-hand side grows faster than the left-hand side (just look at the derivatives). Choosing $y=2x+1$, we can thus deduce
\begin{align*} e^2<\Big(1+\frac{2}{2x+1}\Big)^{2x+2}. \end{align*} Taking the square root, we conclude \begin{align*} e<\Big(1+\frac{2}{2x+1}\Big)^{x+1}. \end{align*}

Edit: Modified due to the oversight pointed out by GFauxPas.

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  • $\begingroup$ Monotone increasing you mean $\endgroup$ – GFauxPas May 30 '19 at 18:12
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    $\begingroup$ @GFauxPas: No, I mean decreasing. I believe $(1+\frac{s}{y})^y$ is increasing and $(1+\frac{s}{y})^{y+1}$ is decreasing towards $e^s$ as $y\rightarrow\infty$ $\endgroup$ – StarBug May 30 '19 at 18:25
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    $\begingroup$ Plotting curves in WA, it seems that whether $(1+s/y)^y$ is increasing or decreasing in $y$ depends on $s$. It seems to be true for $s = 2$ but not $s=3$, so the step needs justification. $\endgroup$ – GFauxPas May 30 '19 at 23:21
  • $\begingroup$ @GFauxPas: Thanks for pointing this out. I guess my original statement was not true for all $s$. However, it is true for $s=2$ (see my modified answer above), so my argument is still valid. $\endgroup$ – StarBug May 31 '19 at 1:24
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    $\begingroup$ @GFauxPas: A computation of the derivative shows that $(1+\frac{s}{y})^{y+1}$ is decreasing in $y$ when $0<s\leq 2$, but not when $s>2$. So I learned something new today:) $\endgroup$ – StarBug May 31 '19 at 2:00
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From $f(x) = (x+1)\log\Big(1 + \frac 2 {2x+1}\Big)$ we compute the derivative.

$$ \begin{eqnarray} f'(x) = & (x+1)'\log\Big(1 + \frac 2 {2x+1}\Big) + (x+1)\log\Big(1 + \frac 2 {2x+1}\Big)' \\ = & (x+1)'\log\Big(1 + \frac 2 {2x+1}\Big) + (x+1)\log\Big(\frac {2x+3} {2x+1}\Big)' \\ = & \log\Big(1 + \frac 2 {2x+1}\Big) + (x+1)\frac {2x+1} {2x+3} \Big( \frac{2(2x+1) - 2(2x+3)}{(2x+1)^2} \Big) \\ = & \log\Big(1 + \frac 2 {2x+1}\Big) + (x+1)\frac {2x+1} {2x+3} \Big( \frac{-4}{(2x+1)^2} \Big) \\ = & \log\Big(1 + \frac 2 {2x+1}\Big) + (x+1)\frac {1} {2x+3} \Big( \frac{-4}{(2x+1)} \Big) \\ = & \log\Big(1 + \frac 2 {2x+1}\Big) - \frac {4(x+1)} {4x^2+8x+3} \end{eqnarray} $$

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  • $\begingroup$ The question was how to show $f'(x) < 0$ on the given interval $\endgroup$ – GFauxPas May 30 '19 at 23:27
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$\quad$The question currently asks for a formal demonstration that

$$\ln\left(1+ \dfrac 2 {2x+1}\right) < \dfrac {4(x+1)}{(4x^2+8x+3)} \text{ for } x\in[1,+\infty).$$

$\quad$To show it's true for $x=1$, we must show $\ln\left(1+\dfrac23\right)<\dfrac{4(2)}{4+8+3}=\dfrac8{15}.$

By exponentiating both sides, this is equivalent to $\dfrac53<\exp\left(\dfrac8{15}\right).$

Well, $\dfrac53=\dfrac{375}{225}<\dfrac{377}{225}=1+\dfrac8{15}+\dfrac12\left(\dfrac{8}{15}\right)^2<\exp\left(\dfrac8{15}\right).$

$\quad$Furthermore, it can be shown that

$$\lim_{x\to\infty}\ln\left(1+ \dfrac 2 {2x+1}\right) = \lim_{x\to\infty}\dfrac {4(x+1)}{(4x^2+8x+3)} =0.$$

$\quad$Finally, it can be shown that $\ln\left(1+ \dfrac 2 {2x+1}\right)$ and $\dfrac {4(x+1)}{(4x^2+8x+3)}$ are monotonically decreasing in the relevant interval. If $0\le x _0<x_1,$ then $2x_0+1<2x_1+1,$ so $\dfrac1{2x_1+1}<\dfrac1{2x_0+1},$

so $\ln\left(1+ \dfrac 2 {2x_1+1}\right) < \ln\left(1+ \dfrac 2 {2x_0+1}\right),$ and $4x_1x_0^2+4x_0^2+5x_0<4x_0x_1^2+4x_1^2+5x_1; $

adding $8x_0x_1+3x_1+3x_0+3$ to both sides,

$4x_1x_0^2+8x_0x_1+3x_1+4x_0^2+8x_0+3<4x_0x_1^2+8x_0x_1+3x_0+4x_1^2+8x_1+3$

i.e., $4(x_1+1)(4x_0^2+8x_0+3)<4(x_0+1)(4x_1^2+8x_1+3),$ so $$\dfrac{4(x_1+1)}{(4x_1^2+8x_1+3)}<\dfrac{4(x_0+1)}{(4x_0^2+8x_0+3)}.$$

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