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It is well known that for $x>0$ that $\left(1+\frac{1}{x}\right)^x\le e\le\left(1+\frac{1}{x}\right)^{x+1}$ (see wikipedia). However, one can obtain the stronger inequality $$ \sqrt{1+\frac{1}{x+\frac{1}{2}}}\left(1+\frac{1}{x}\right)^x\le e\le\sqrt{1+\frac{1}{x}}\left(1+\frac{1}{x}\right)^{x} $$ The second inequality can be found in Proposition B.3 of "Randomized Algorithms", by Raghaven and Motwani (which itself refers to the book "Analytic Inequalities" by Mitrinović) , and can be proven straight-forwardly by calculus (showing a first derivative is non-negative and such).

While I can also prove the first inequality using familiar calculus methods, it is a bit messy (ultimately requiring that $\frac{1}{y+2}+\frac{1}{3y+2}\ge \frac{1}{y+1}$ for $y\ge 0$).

Does anyone know a "simple" proof of $\sqrt{1+\frac{1}{x+\frac{1}{2}}}\left(1+\frac{1}{x}\right)^x\le e$?

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    $\begingroup$ The title doesn't accurately describe the question asked. $\endgroup$ Dec 29 '15 at 5:33
  • $\begingroup$ Should the exponent on the right hand side perhaps also be $x$ instead of $x+1$? As stated the inequality is weaker than the "standard" one. $\endgroup$
    – WimC
    Dec 29 '15 at 10:25
  • $\begingroup$ I didn't fully understand the question, is a proof using derivative accepted? $\endgroup$
    – mrprottolo
    Dec 29 '15 at 11:45
  • $\begingroup$ I think the right most expression of the second inequality is larger than the right hand expression of the first. So the second inequality isn't strictly stronger. $\endgroup$
    – Joel
    Dec 29 '15 at 14:52
  • $\begingroup$ @GregMartin: I updated the title. Hopefully this is more clear. $\endgroup$
    – miforbes
    Dec 29 '15 at 21:50
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You can do better than the stated inequality starting from $$\log(1+x)<\frac{x(x+6)}{4x+6}$$ for all $x>0$. (Equality for $x=0$ and the difference between the right- and left hand side is strictly increasing as can be shown by taking derivatives.) With this we find $$ \frac{1}{2}\log\left(1+\frac{1}{x+\frac{1}{3}}\right)+x \log\left(1+\frac{1}{x}\right) < \frac38\left(\frac1{3x+1}+\frac1{x+1}\right)+\left(1-\frac3{6x+4}\right)=$$ $$1-\frac{3x}{4(3x+1)(x+1)(3x+2)}<1$$ for all $x>0$. In fact one can show in the same way that for any $\alpha > \tfrac16$ the inequality $$ \frac{1}{2}\log\left(1+\frac{1}{x+\alpha}\right)+x \log\left(1+\frac{1}{x}\right) < 1$$ holds for all $$x>\max\left(0,\frac{2-18\alpha^2}{18\alpha -3}\right).$$

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  • $\begingroup$ Thanks. Heuristically I believe that $\alpha=\frac{1}{e^2-1}$ might work (for all $x$), but can't prove it, but I only really care about $\alpha=1/2$ here. $\endgroup$
    – miforbes
    Dec 29 '15 at 21:56
  • $\begingroup$ Unfortunately the proof you give seems about as involved as the one I came up with. Admittedly this is not too difficult, but it seems more annoying that the proof of the above RHS of the main inequality, which is simpler. $\endgroup$
    – miforbes
    Dec 29 '15 at 21:58
  • $\begingroup$ @miforbes Yes it is not exactly a one liner. It seems that this inequality (with $\alpha=\tfrac12$) is quite strong. I would be interested in your method (or any other for that matter). $\endgroup$
    – WimC
    Dec 30 '15 at 7:21
  • $\begingroup$ I posted my proof, but I'm not sure it adds any insight here. $\endgroup$
    – miforbes
    Dec 31 '15 at 4:22
  • $\begingroup$ Initially the inequality $\log(1+x)<\frac{x(x+6)}{4x+6}$ you gave seemed arbitrary, but I found out that it is a Pade approximation ("Some bounds for the logarithmic function", Topsoe), which makes it more motivated. $\endgroup$
    – miforbes
    Aug 11 '16 at 21:21
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This does not prove the inequality for every $x$, but only when $x$ is succiently large. Taking the $\log$ of the inequality: $$\frac{1}{2}\log \Big(\frac{1}{x+\frac{1}{2}}+1 \Big)+x \log \Big(\frac{1}{x}+1 \Big) -1<0.$$ Using series expansion for $x=\infty$: \begin{align} & \frac{1}{2x+1}-\frac{1}{(2x+1)^2} +x\Big(\frac{1}{x}-\frac{1}{2x^2}+\frac{1}{3x^3}\Big)+-1+o(\frac{1}{x^2})= \\ &=\frac{-4x^2+5x+2}{6(2x+1)^2x^2}+o(\frac{1}{x^2}) \end{align} Hence the last quantity is less than $0$ for $x$ sufficiently large, and this implies that your inequality is true when $x$ is big enaugh.

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  • $\begingroup$ Thanks, I suppose I'm looking for a proof that works for all $x$. $\endgroup$
    – miforbes
    Dec 29 '15 at 21:54
  • $\begingroup$ You miss a term $1/(3x^2)$ from the second $\log$ expression. $\endgroup$
    – WimC
    Dec 30 '15 at 7:31
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I found a proof that seems well-motivated enough to satisfy me. The key lemma is the Hermite-Hadamard inequality (see wikipedia, or Old and new on the Hermite-Hadamard inequality).

Lemma (Hermite-Hadamard): Let $f:[a,b]\to\mathbb{R}$ be convex. Then $$ f\left(\frac{a+b}{2}\right) \le\frac{1}{b-a}\int_a^b f(x)dx \le \frac{f(a)+f(b)}{2}. $$

This can be seen as an instantiation of the trapezoid rule for approximating integrals. The lower bound is Jensen's inequality, the upper bound follows from seeing (via convexity) that the curve $(x,f(x))$ is below the line $(a,f(a))\to(b,f(b))$ on the interval $[a,b]$. That Hermite-Hadamard is useful for the desired inequalities is in retrospect not surprising, as these inequalities can be used to prove Stirling's formula (up to the exact constant), and some proofs of Stirling's formula use Hermite-Hadamard/trapezoid-rule.

I'll first give a weaker result using Hermite-Hadamard which is pretty straightforward, then a proof of the exact result I wanted.


Applying Hermite-Hadamard to $\frac{1}{x}$ on $[x,x+1]$, and simplifying, we get that

$$ \frac{1}{x+\frac{1}{2}} \le \ln\left(1+\frac{1}{x}\right) \le \frac{1}{2}\left(\frac{1}{x}+\frac{1}{x+1}\right) $$

Manipulating the lower bound, we see that

$$1\le \left(x+\frac{1}{2}\right)\ln\left(1+\frac{1}{x}\right)$$

which is equivalent to

$$e\le \sqrt{1+\frac{1}{x}}\left(1+\frac{1}{x}\right)^x.$$

Turning to the other direction, we want to show

$$ \sqrt{1+\frac{1}{x+1}}\left(1+\frac{1}{x}\right)^x\le e $$

Taking logarithms,

$$ \frac{1}{2}\ln\left(1+\frac{1}{x+1}\right)+x\ln\left(1+\frac{1}{x}\right)\le 1 $$

Using that $\ln(1+y)\le y$, and the above upper bound via Hermite-Hadamard, we get

$$ \frac{1}{2}\ln\left(1+\frac{1}{x+1}\right)+x\ln\left(1+\frac{1}{x}\right) \le \frac{1}{2}\cdot\frac{1}{x+1}+x\cdot \frac{1}{2}\left(\frac{1}{x}+\frac{1}{x+1}\right) =\frac{1}{2}\left(\frac{x}{x}+\frac{x+1}{x+1}\right)=1 ,$$

as desired.


To get the tighter bound, we can apply Hermite-Hadamard to $\ln(x)$ (which is concave, so the inequality is reversed) on $[x,x+1]$, so that (after integrating),

$$ \frac{1}{2}(\ln(x+1)+\ln(x)) \le \ln(x+1)+x\ln\left(1+\frac{1}{x}\right)-1 \le \ln\left(x+\frac{1}{2}\right) $$

Manipulating the upper bound yields

$$ 1 \ge x\ln\left(1+\frac{1}{x}\right)+\ln\left(\frac{x+1}{x+\frac{1}{2}}\right) =x\ln\left(1+\frac{1}{x}\right)+\ln\left(1+\frac{1}{2}\cdot\frac{1}{x+\frac{1}{2}}\right). $$

Exponentiating, this is

$$ e \ge \left(1+\frac{1}{2}\cdot \frac{1}{x+\frac{1}{2}}\right)\left(1+\frac{1}{x}\right)^x $$

Appealing to Bernoulli's inequality, $1+y/2\ge \sqrt{1+y}$

$$ e\ge \sqrt{1+\frac{1}{x+\frac{1}{2}}}\left(1+\frac{1}{x}\right)^x, $$

as desired. (One can also use the lower bound for Hermite-Hadamard on $\ln(x)$ to again get $e\le (1+1/x)^{x+1/2}$.)

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In response to @WimC, I'm posting my own proof of this inequality. The first observation is to see that $\sqrt{1+\frac{1}{x+1/2}}(1+1/x)^x\le e$ for $x\ge 0$ is equivalent to (under $y:=1/x$), $$\left(1+\frac{2y}{y+2}\right)^{y/2}(1+y)\le e^y$$ for $y>0$. One can then hope to establish this inequality for all $y \ge 0$.

Taking logarithms, this reduces to showing that $$f(y):=y-\ln(y+1)-\frac{y}{2}\ln(3y+2)+\frac{y}{2}\ln(y+2)$$ has $f(y)\ge 0$ for $y\ge 0$. As $f(0)=0$, one needs only show the derivative $f'(y)\ge 0$ for $y\ge 0$. Taking the derivative, one finds that \begin{align*} f'(y)&=1-\frac{1}{y+1}-\frac{1}{2}\ln\left(1+\frac{2y}{y+2}\right)+\frac{1}{2}\cdot\frac{y}{y+2}-\frac{1}{2}\cdot \frac{3y}{3y+2}\\ &\ge 1-\frac{1}{y+1}-\frac{1}{2}\cdot \frac{2y}{y+2}+\frac{1}{2}\cdot\frac{y}{y+2}-\frac{1}{2}\cdot \frac{3y}{3y+2}\\ &=\frac{y^2}{(y+1)(y+2)(3y+2)} \end{align*} where we have used $\ln(1+z)\le z$, then simplified. This last expression is clearly non-negative, as desired.


My issue with the above proof is that the algebra is messy, and I don't gain any intuition on why the inequality is true.


In particular, I would ideally like to be able to get the best $\alpha$ so that $$\sqrt{1+\frac{1}{x+1/2}}(1+1/x)^x\le e$$ is true for all $x>0$. Using the above $y:=1/x$, this is equivalent to $$\sqrt{1+\frac{1}{1/y+\alpha}}(1+y)^{1/y}\le e$$ If we take $y\rightarrow \infty$ then as $1+y\sim e^{\ln y}$ we get $(1+y)^{1/y}\rightarrow 1$, so that we must have that for large $y$ that $$\sqrt{1+\frac{1}{0+\alpha}}\le e$$ Rearranging, this says that $\alpha$ must obey $\alpha\ge\frac{1}{e^2-1}\approx\frac{1}{6.39}$ for large $y$. To me it seems natural to expect that any such $\alpha$ works for all $y$, but that might be naive.

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  • $\begingroup$ Thanks for taking the time to write this down. Since $$\tfrac12\log\left(1+\frac1{x+\alpha}\right)+x \log\left(1+\frac1x\right)=1+\frac{1-6\alpha}{12}x^{-2}+o(x^{-2})$$ we can conclude that $\alpha$ must be at least $\frac16$ so $(e^2-1)^{-1}$ will not work. $\endgroup$
    – WimC
    Dec 31 '15 at 7:19

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