# Prove that $\sqrt{7}^{\sqrt{8}}>\sqrt{8}^{\sqrt{7}}$

show that $$\sqrt{7}^{\sqrt{8}}>\sqrt{8}^{\sqrt{7}}$$

and I found $$LHs-RHS=0.017\cdots$$

I have post this interesting problem Prove $\left(\frac{2}{5}\right)^{\frac{2}{5}}<\ln{2}$

can someone suggest any other nice method? Thank you everyone.

• It is true that $7^8>8^7$, since $(1+\frac{1}{7})^7<e<7$. Then we have, $\sqrt{7}^8>\sqrt{8}^7$. Does it imply that $\sqrt{7}^\sqrt{8}>\sqrt{8}^\sqrt{7}$. – mtm Jun 28 '13 at 9:26
• No,How can have $\sqrt{7}^8>\sqrt{8}^7$? – math110 Jun 28 '13 at 9:31
• $\sqrt{7}^{16}>\sqrt{8}^{14}$. So if we take $\frac{1}{2}$ power of both sides, we have $\sqrt{7}^{8}>\sqrt{8}^{7}$. Maybe i am wrong. – mtm Jun 28 '13 at 9:37
• $\sqrt{7}^{8}>\sqrt{8}^{7}\Longrightarrow \sqrt{7}^{\sqrt{8}}>\sqrt{8}^{\sqrt{7}}$? I think is wrong. – math110 Jun 28 '13 at 10:58
• @mtm Monotonicity argument like yours will not work here, no matter you take $(\ln t)/\sqrt{t}$ or $(\ln t)/t$. – Shuhao Cao Jun 28 '13 at 15:21

Note that $\sqrt{7}^{\sqrt{8}}\doteq15.673$ and $\sqrt{8}^{\sqrt{7}}\doteq15.656$, so these two are pretty close. Furthermore $\sqrt{7}<e<\sqrt{8}$, and the function $f(x):={\log x\over x}$ has a local maximum at $x:=e$. This excludes the use of monotonicity arguments. The following proof uses integer arithmetic instead.

To begin with we need a rational approximation to $\sqrt{7\over8}$ that is slightly larger than $\sqrt{7\over8}$. Using Mathematica (or continued fractions) one finds that $$7\cdot 31^2=6727<6728=8\cdot 29^2\ ,$$ which implies $$\sqrt{7}\cdot31<29\cdot\sqrt{8}\ .$$ Furthermore one computes $$8^{29}=15\>47425\>04910\>67253\>43623\>90528<15\>77753\>82034\>84580\>66150\>42743=7^{31}\ .$$ It follows that $$\bigl(8^{\sqrt{7}}\bigr)^{31}<\bigl(8^{29}\bigr)^{\sqrt{8}}<\bigl(7^{31}\bigr)^{\sqrt{8}}\ .$$ Now take the $62^{\rm th}$ root on both sides.

• Very instructive! +1 – Markus Scheuer Nov 13 '14 at 7:58

The challenge here is to demonstrate the inequality in a way that can easily be checked by hand. Taking logs, the inequality to be proved is

$$\sqrt7\ln8 \lt \sqrt8\ln7$$

which can be written as

$$3\sqrt{1-{1\over8}}\ln2 \lt \ln(8-1) = 3\ln2+\ln(1-{1\over8})$$

so the inequality to prove is

$$-\ln\left(1-{1\over8}\right)\lt 3\left(1-\sqrt{1-{1\over8}}\right)\ln2$$

We'll use the Taylor series

$$-\ln(1-x)=x+{1\over2}x^2+{1\over3}x^3+{1\over4}x^4+\cdots$$

to obtain

\begin{align} -\ln\left(1-{1\over8}\right) &\lt {1\over8}+{1\over2}\left({1\over8}\right)^2+{1\over3}\left({1\over8}\right)^3+{1\over3}\left({1\over8}\right)^4\\ &={1\over8}+{1\over2}\left({1\over8}\right)^2+{1\over3}\left({1\over8}\right)^3{1\over1-{1\over8}}\\ &={1\over8}+{1\over128}+{1\over21}{1\over64}\\ &\lt{1\over8}+{1\over128}+{1\over16}{1\over64}\\ &={128+8+1\over2^{10}}\\ &={137\over2^{10}} \end{align}

We'll use the fact that all the coefficients in the Taylor series

$$1-\sqrt{1-x}={1\over2}x+{1\over8}x^2+{1\over16}x^3+\ldots$$

are positive to get

$$1-\sqrt{1-{1\over8}}\gt {1\over2}{1\over8}+{1\over8}\left({1\over8}\right)^2={33\over2^9}$$

Finally, we use the easily verified identity

$$\ln2=3\ln\left(1+{1\over4}\right)+\ln\left(1+{3\over125}\right)$$

and the alternating Taylor series

$$\ln(1+x)=x-{1\over2}x^2+{1\over3}x^3-\cdots$$

to get

\begin{align} \ln2&\gt 3\left({1\over4}-{1\over2}\left({1\over4}\right)^2+{1\over3}\left({1\over4}\right)^3-{1\over4}\left({1\over4}\right)^4\right)+{3\over125}\\ &={21\over32}+{1\over64}-{3\over1024}+{3\over125}\\ &\gt{21\over32}+{1\over64}-{3\over1024}+{3\over128}\\ &={21\cdot32+16-3+24\over1024}={709\over1024} \end{align}

hence

$$3\ln2\gt{2127\over1024}\gt{2126\over1024}={1063\over512}$$

so that

$$3\left(1-\sqrt{1-{1\over8}}\right)\ln2\gt{33\over2^9}{1063\over2^9}={33\cdot1063\over2^{18}}$$

Consequently, the inequality to verify is

$${137\over2^{10}}\lt{33\cdot1063\over2^{18}}$$

which is to say,

$$256\cdot137\lt33\cdot1063$$

Eschewing calculators, note that $1063=1024+32+8-1$, so

$$33\cdot1063=(2^5+1)(2^{10}+2^5+2^3-1)=2^{15}+2^{11}+2^8+2^3-1$$

whereas

$$256\cdot137=2^8(2^7+2^3+1)=2^{15}+2^{11}+2^8$$

so the desired inequality holds.

• Reading along, I kept expecting the computation to end, but it never did! A rather intimidating verification. – Ryan Reich Aug 8 '13 at 5:29
• @RyanReich, but it did end! ;-) – Barry Cipra Aug 8 '13 at 6:00

I just want to give you a bigger picture on this question. The fact you point out corresponds to $f(2)>0.$

$f(\text{x$\_$})\text{:=}\left(t=\lfloor \exp (x)\rfloor ;a=t^{1/x};b=(t+1)^{1/x};a^b-b^a\right);$

$\text{Plot}[f(x),\{x,1,3\}]$