simple proof that $\sqrt{1+\frac{1}{x+1/2}}(1+1/x)^x\le e$

Question

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

Answer 1

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.

Answer 2

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 $$x>\max\left(0,\frac{2-18\alpha^2}{18\alpha -3}\right).$$

Answer 3

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.

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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.

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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.