# Which number is greater, $2^\sqrt2$ or $e$?

Claim: $\color{red}{2^\sqrt2<e}$

Note: $2^\sqrt2=e^{\sqrt2\ln2}$

Different approach: We show $e^{x-1}>x^\sqrt x$ for $x>2$.

Let $f(x) = x -1 - \sqrt{x} \ln x$. We have $f'(x) = 1 - \frac{ \ln x }{2 \sqrt{x} } - \frac{1}{\sqrt{x}}$. By inspection, note that $f'(1)=0$ and since for $x>1$, pick $x=4$, for instance, we have $f'(4)=1- \frac{ \ln 4 }{4} - \frac{1}{2} = \frac{1}{2} - \frac{ \ln 2 }2 > 2$, then we know $f(x)$ is increasing for $x> 1$, Thus

$$f(x) > f(1) \implies x - 1 - \sqrt{x} \ln x > 0 \implies x - 1 \geq \sqrt{x} \ln x \implies e^{x-1} > x^{\sqrt{x}}$$

Putting $x=2$, we obtain $$e>2^\sqrt2$$ What do you think about this approach? Is there an easier way?

• WWND: (What would Newton do?) Take logs and numerically estimate the left hand side by a Taylor series until you can conclude. Commented Oct 23, 2016 at 6:07
• is very good, I think, something elegant. Commented Oct 29, 2016 at 12:46

To show $2^{\sqrt{2}}<e$, it suffices to show that $\sqrt{2}\ln 2 < 1$, or $\ln 2 < \frac{1}{\sqrt{2}}$. Notice that $\ln 2 = \int\limits_{1}^{2}{\frac{1}{x}\,dx}$, and by Cauchy-Schwarz we have $$\ln 2 = \int\limits_{1}^{2}{\frac{1}{x}\cdot 1\,dx}\le\left(\int\limits_{1}^{2}{\frac{1}{x^2}\,dx}\right)^{1/2}\left(\int\limits_{1}^{2}{1\,dx}\right)^{1/2} = \left(\frac{1}{2}\right)^{1/2}\left(1\right)^{1/2} = \frac{1}{\sqrt{2}}.$$ Equality cannot hold as $(1/x)/1$ is not constant, so it follows that $\ln 2 < \frac{1}{\sqrt{2}}$, as desired.

• The best point in your proof it's that the equality does not occur. :D Commented Oct 23, 2016 at 6:53

The inequality $e>2^\sqrt2$ is equivalent to $$e^{\sqrt{2}}>(2^\sqrt2)^{\sqrt{2}}=2^2=4.$$ Now $$e^{\sqrt{2}}>\sum_{k=0}^5\frac{(\sqrt{2})^k}{k!}=\frac{13}{6}+\frac{41\sqrt{2}}{30}>\frac{13}{6}+\frac{4}{3}\cdot\frac{7}{5}=\frac{121}{30}>4$$ because $\sqrt{2}>7/5$ (equivalent to $2\cdot 5^2>7^2$).

Approximating the integral by the area of a trapezoid we get for $x>1$ $$\log(x)=\int_1^x\frac{\mathrm{d}t}{t}<(x-1)\frac{1+\frac1x}2=\frac{x^2-1}{2x}$$ So $$\sqrt{2}\log(2)=2\sqrt{2}\log(\sqrt{2})<1.$$

I have an approach take logs we get $1,\sqrt {2 }ln (2)$ then using series expansion of $\ln (\frac{1+x}{1-x})$ for $ln (2)$ we have $x=1/3$ the series is $2 (x+\frac {x^3}{3} ...)$ we get $\sqrt {2}ln (2) =1.4(2)(\frac {1}{3}+\frac {1}{81} ) <1$ thus $e>2^{\sqrt {2} }$

This is a question involving numerical computations.

Here is an attempt. Consider the unproved inequality $e>2^{\surd 2}$. By raising to the power $\surd2$ on both sides the above inequality would be true iff $e^{\surd 2}> {(2^{\surd2}})^{\surd2}= 4$.

Now taking 8 terms of the series for $e^x$ and putting $x=\surd2$ (separating even and odd terms).

$$e^{\surd2}> \big(1+\frac22 + \frac4{4!} + \frac8{6!}\big) + \surd2\big(\frac2{3!}+ \frac4{5!}+\frac8{7!}\big) .$$ Now the rhs is seen to be bigger than 4 when we substitue 1.414 for $\surd2$.

Another way:

since $e^x$ is increasing and $\sqrt2\approx1.414213$ we have $$(2^{\sqrt2})^{\sqrt2}=4\lt e^{1.4}\approx4.055199\lt e^{\sqrt2}$$It follows $$\left((2^{\sqrt2})^{\sqrt2}\right)^{\frac{1}{\sqrt2}}\lt(e^{\sqrt2})^{\frac{1}{\sqrt2}}\iff2^{\sqrt2}\lt e$$