# Prove that $\left(1+2^{-1+b^{\left(\frac{1}{b-1}\right)}}\right)^b < 1+2^{-1+b^{\left(\frac{b}{b-1}\right)}}$ for all $b>2$.

While solving a bigger problem, I've reduced it to an inequality $$\left(1+2^{-1+b^{\left(\frac{1}{b-1}\right)}}\right)^b < 1+2^{-1+b^{\left(\frac{\color{red}b}{b-1}\right)}}$$for $$b>2$$, which looks plausible when looking at the plots. I've tried using Jensen inequality here with $$x\mapsto x^{b-1}$$, but not much luck.

I've also checked that the inequality works empirically with Wolfram Alpha.

Yes, this is written correctly, opposite to what some of the people in the comments below are trying to claim. I'm surprised this even needs claryfying.

• What are you trying to do? Do you want to solve the inequality for b? Jul 5, 2020 at 17:29
• I agree with @RadialArmSaw , do you want a proof that it is correct for every $b>2$? Moreover, the exponents of the $2$ are not very clear, it seems something is misplaced. Jul 5, 2020 at 18:03
• Yes, proof that the inequality holds for $b>2$. The inequality is written correctly. Jul 5, 2020 at 18:13
• Would you mind including what the "bigger problem" was in your question? Jul 5, 2020 at 21:23
• Ok that's alright. Maybe one approach to the problem would be showing that $\frac{d}{db}($RHS - LHS$) > 0$ for $b \ge 2$? I don't know how feasible this is though. Jul 5, 2020 at 21:46

Case $$b > \frac{7}{2}$$:

It suffices to prove that $$2^{-1 + b^{\frac{b}{b-1}}} \ge \left(1 + 2^{-1 + b^{\frac{1}{b-1}}}\right)^b$$ or $$2^{-\frac{1}{b} + b^{\frac{1}{b-1}}} \ge 1 + 2^{-1 + b^{\frac{1}{b-1}}}$$ or $$2^{b^{\frac{1}{b-1}}}\left(2^{-\frac{1}{b}} - 2^{-1}\right) \ge 1.$$

To proceed, we need the following results. The proofs are given at the end.

Fact 1: $$2^{-\frac{1}{b}} > \frac{21b+4}{21b+21}$$ for $$b > \frac{7}{2}$$.

Fact 2: $$b^{\frac{1}{b-1}} > \frac{13b + 51}{13b + 13}$$ for $$b > \frac{7}{2}$$.

Let us proceed. By Facts 1 and 2, it suffices to prove that $$2^{\frac{13b + 51}{13b + 13}}\left(\frac{21b+4}{21b+21} - 2^{-1}\right) \ge 1$$ or $$\frac{13b + 51}{13b + 13} \ln 2 + \ln \left(\frac{21b+4}{21b+21} - 2^{-1}\right) \ge 0.$$ Denote LHS by $$F(b)$$. We have $$F'(b) = \frac{(-798\ln 2 + 442)b+494\ln 2+442}{13(21b-13)(b+1)^2}$$. From $$F'(b)=0$$, we have $$b = \frac{494\ln 2+442}{798\ln 2 - 442} \approx 7.058440304$$. Clearly, $$F(b)$$ is strictly increasing on $$b \in (\frac{7}{2}, \frac{494\ln 2+442}{798\ln 2 - 442})$$, and strictly deceasing on $$b \in (\frac{494\ln 2+442}{798\ln 2 - 442}, \infty)$$. Also, $$F(\frac{7}{2}) > 0$$ and $$F(\infty) = 0$$. Thus, $$F(b) > 0$$ on $$(\frac{7}{2}, \infty)$$. We are done.

$$\phantom{2}$$

Proof of Fact 1: It suffices to prove that $$-\frac{1}{b} \ln 2 - \ln \frac{21b+4}{21b+21} > 0.$$ Denote LHS by $$f(b)$$. We have $$f'(b) = \frac{(21\ln 2-17)b^2 + 25b\ln 2+4\ln 2}{(21b+4)(b+1)b^2}$$. It is easy to prove that $$f(b)$$ is strictly increasing on $$(\frac{7}{2}, b_0)$$, and strictly decreasing on $$(b_0, \infty)$$, where $$b_0 = \frac{25\ln 2 + \sqrt{289(\ln 2)^2 + 272\ln 2}}{34 - 42 \ln 2} \approx 7.247101660$$. Also, $$f(\frac{7}{2}) > 0$$ and $$f(\infty) = 0$$. Thus, $$f(b) > 0$$ on $$b\in (\frac{7}{2}, \infty)$$. We are done.

$$\phantom{2}$$

Proof of Fact 2: It suffices to prove that $$\ln b - (b-1)\ln \frac{13b + 51}{13b + 13} > 0.$$ Denote LHS by $$g(b)$$. We have \begin{align} g'(b) &= \frac{51b^2+26b+51}{(13b+51)(b+1)b} - \ln \frac{13b + 51}{13b + 13}\\ &= \frac{51b^2+26b+51}{(13b+51)(b+1)b} - \ln\left(1 + \frac{38}{13b + 13}\right) \\ &> \frac{51b^2+26b+51}{(13b+51)(b+1)b} - \frac{38(2535b^2+10257b+8444)}{39(845b^2+4654b+5975)(b+1)}\\ &= \frac{428415b^4+133848b^3-5765266b^2-1049016b+11884275}{39(13b+51)(b+1)b(845b^2+4654b+5975)}\\ &> 0 \end{align} where we have used $$\ln (1+x) < \frac{x(x^2+21x+30)}{3(3x^2+12x+10)}$$ for $$x > 0$$ (the proof is easy), and let $$b = \frac{7}{2} + y$$ for $$y > 0$$ to get \begin{align} &428415b^4+133848b^3-5765266b^2-1049016b+11884275\\ =\ & \frac{121855511}{16} + \frac{73972417}{2}y + \frac{54257281}{2}y^2+ 6131658y^3 + 428415y^4\\ > \ & 0. \end{align} Also, $$g(\frac{7}{2}) > 0$$. Thus, we have $$g(b) > 0$$ on $$(\frac{7}{2}, \infty)$$. We are done.

The result follows from Proposition 1 and Proposition 2 below.

Lemma 1. $$\ln(2)\cdot (b^{\frac 1 {b-1}}-1)>\frac 2 {b-1}$$ if $$b>e^{2/\ln(2)}\approx 17.91.$$

Proof. $$\ln(2)\cdot (b^{\frac 1 {b-1}}-1)-\frac 2 {b-1}$$ $$=\ln(2)(e^{\frac 1{b-1}\ln(b)}-1)-\frac 2{b-1}$$ $$\geq \ln(2)(1+\frac {\ln(b)}{b-1}-1)-\frac 2{b-1}$$ $$=\frac 1{b-1}(\ln(2)\cdot \ln(b)-2),$$ which is positive if $$\ln(2)\cdot \ln(b)-2>0\Leftrightarrow b>e^{2/\ln(2)}.$$

Lemma 2. $$2^{b^{\frac 1{b-1}}-1}>1+\frac 2{b-1}$$ if $$b>e^{2/\ln(2)}$$.

Proof. Assume $$b>e^{2/\ln(2)}$$. Then $$2^{b^{\frac 1{b-1}}-1}=e^{\ln(2)(b^{\frac 1{b-1}}-1)}\geq 1+\ln(2)(b^{\frac 1{b-1}}-1)>1+\frac 2{b-1},$$ by Lemma 1.

Proposition 1. $$\left(1+2^{b^{\frac 1{b-1}}-1}\right)^b<1+ 2^{b^{\frac b{b-1}}-1}$$ if $$b>e^{2/\ln(2)}.$$

Proof. Let $$2^{b^{\frac 1{b-1}}-1}=1+\epsilon,$$ where $$\epsilon>0$$ depends on $$b$$. Then the given statement is equivalent to $$(2+\epsilon)^b<1+\frac{(2+2\epsilon)^b}2,\quad (1)$$ as is easily checked (note that $$2^{b^{\frac b{b-1}}-1}=\frac 1 2\cdot 2^{b^{\frac b{b-1}}}=\frac 1 2\cdot 2^{b^{1+\frac 1{b-1}}}=\frac 1 2\cdot 2^{b^{\frac 1{b-1}}\cdot b})$$.

To show (1), one proves the following stronger result which implies (1): $$\left(1+\frac {\epsilon}{2+\epsilon}\right)^b=\left(\frac{2+2\epsilon}{2+\epsilon}\right)^b>2.\quad (2)$$

Since $$(1+x)^b\geq 1+bx$$ for all $$b\geq 1,x\geq 0$$, one has $$\left(1+\frac {\epsilon}{2+\epsilon}\right)^b\geq 1+\frac{b\epsilon}{2+\epsilon},$$ so to prove (2), it suffices to show that $$\frac{b\epsilon}{2+\epsilon}>1$$ Using the definition of $$\epsilon$$, this is equivalent to $$b\epsilon>2+\epsilon$$ $$\Leftrightarrow b\left(2^{b^{\frac 1{b-1}}-1}-1\right)>2+2^{b^{\frac 1{b-1}}-1}-1$$ $$\Leftrightarrow 2^{b^{\frac 1{b-1}}-1}>1+\frac 2{b-1},$$ so the result follows from Lemma 2. QED

Now given Proposition 1, in order to prove the result in the original post, it suffices to show the following.

Proposition 2. The function $$f(b):=1+ 2^{b^{\frac b{b-1}}-1}-\left(1+2^{b^{\frac 1{b-1}}-1}\right)^b>0$$ for all $$b\in (2,18]$$

Proof.

Step 1. By numerical method, one can show that $$f(b)>0$$ for all $$b\in [10003/5000,18].$$ The idea is to subdivide the interval $$[2,18]$$ into $$N=2000000$$ equal parts and estimate the value of $$f(b)$$ on each interval $$I_i:=[x_{i-1},x_i]$$, where $$i=1,\cdots,N$$ and $$x_i-x_{i-1}=(18-2)/N$$. To do this, one estimates the lower bound of $$f(b)$$ on $$I_i$$, namely $$f(b)\geq 1+2^{x_{i-1}^{\frac{x_{i-1}}{x_{i-1}-1}}-1}-\left(1+2^{x_{i-1}^{\frac 1{x_{i-1}-1}}-1}\right)^{x_{i}},$$ where one uses the fact that $$b^{\frac b{b-1}}$$ is increasing and $$b^{\frac 1{b-1}}$$ is decreasing (for $$b>2$$). The data from SAGE shows that $$f(b)$$ is positive on $$I_i$$ for $$i\geq 76$$, which means that $$f(b)>0$$ for $$b\in [10003/5000,18].$$

Step 2. Show that $$f(b)>0$$ for $$b\in (2,10003/5000)$$. Since $$f(2)=0$$, one just needs to show that $$f'(2)>0$$ and $$f'(x)>0$$ on a small neighborhood containing $$(2,10003/5000)$$. Again this can be done by similar method as in Step 1 for $$f'(x)$$. As this will not be difficult, details are omitted.

This is a partial answer, at least. I'm planning to keep thinking about this tomorrow, but maybe someone else can step in and finish what I have. The argument is almost complete, but I have other things to work on right now.

First we log both sides to get the equivalent inequality

$$b \log \left ( 1+2^{b^{\frac{1}{b-1}}-1} \right ) \overset{?}{<} \log \left ( 1+2^{b^{\frac{b}{b-1}}-1} \right )$$

Then we factor out the dominant term of each log, and separate to get

$$b \left [ \log \left ( 2^{b^{\frac{1}{b-1}}-1} \right ) + \log \left ( 1 + 2^{1 - b^{\frac{1}{b-1}}} \right ) \right ] \overset{?}{<} \log \left ( 2^{b^{\frac{b}{b-1}}-1} \right ) + \log \left ( 1 + 2^{1 - b^{\frac{b}{b-1}}} \right )$$

Then we apply some log rules and rearrange

$$b^{\frac{b}{b-1}} \log(2) - b \log(2) + b \log \left ( 1 + 2^{1 - b^{\frac{1}{b-1}}} \right ) \overset{?}{<} b^{\frac{b}{b-1}} \log(2) - \log(2) + \log \left ( 1 + 2^{1 - b^{\frac{b}{b-1}}} \right )$$

We can cancel the first term of each side, and swap the second terms to make them positive

$$\log(2) + b \log \left ( 1 + 2^{1 - b^{\frac{1}{b-1}}} \right ) \overset{?}{<} b \log(2) + \log \left ( 1 + 2^{1 - b^{\frac{b}{b-1}}} \right )$$

Now, looking at the left hand side, notice $$b^{\frac{1}{b-1}} \to 1$$ from above. So we get the following honest upper bound on the left hand side

$$\log(2) + b \log \left ( 1 + 2^{1 - b^{\frac{1}{b-1}}} \right ) < \log(2) + b \log(2)$$

In fact, this inequality is not very tight -- It turns out for $$b > 4$$, we have

$$\log(2) + b \log \left ( 1 + 2^{1 - b^{\frac{1}{b-1}}} \right ) < b \log(2)$$

but my only proof (so far) is desmos:

Looking at the right hand side, notice $$b^{\frac{b}{b-1}} \to \infty$$, so $$2^{1-b^{\frac{b}{b-1}}} \to 2^{1-\infty} \to 0$$, so it's a good thing that $$b \log(2)$$ eventually dominates the left hand side!

In summary, by actually proving the tighter inequality $$\log(2) + b \log \left ( 1 + 2^{1 - b^{\frac{1}{b-1}}} \right ) < b \log(2)$$, which I think is a reasonable goal, we can show that your desired inequality holds for $$b > 4$$ (really $$3.384$$).

But we're currently using $$0$$ as our lower bound for $$\log \left ( 1 + 2^{1-b^{\frac{b}{b-1}}} \right )$$. One can see from the graph that the desired inequality holds from $$2$$ to $$3.384$$, so if that pleases you then we're done. If not, then slightly more work is needed, but I haven't spent much time thinking about this case yet.

I hope this helps ^_^

My new proof (sketch)

Let $$b > 2$$. Prove that $$1 + 2^{-1 + b^{\frac{b}{b-1}}} \ge \left(1 + 2^{-1 + b^{\frac{1}{b-1}}}\right)^b.$$

We first give the following auxiliary results (Facts 1-4).

Fact 1: $$2^{b^{\frac{1}{b-1}}} \ge \frac{30b+52}{15b-2}$$ for all $$b \in [2, \infty)$$.

Fact 2: Let $$b_1 > 2$$ be given. Let $$p(x) = 1 + 2^{-1} x^{b_1} - (1 + 2^{-1} x)^{b_1}$$. Then $$p(x)$$ is strictly increasing on $$(2, \infty)$$. (Hint: $$p'(x) = \frac{1}{2} b_1 \left(x^{b_1-1} - \left(1+\frac{x}{2}\right)^{b_1-1}\right) > 0$$. )

Fact 3: $$\left(\frac{9}{16}\right)^{2/b} \ge \frac{363b - 168}{360b + 272}$$ for all $$2\le b \le 4$$.

Fact 4: $$9^{2/b} \ge \frac{(29b+50)^2}{36(5b-4)^2}$$ for all $$2\le b \le 4$$.

Now, the desired inequality is written as $$1 + 2^{-1}\left(2^{b^{\frac{1}{b-1}}}\right)^b \ge \left(1 + 2^{-1}2^{b^{\frac{1}{b-1}}}\right)^b.$$ By Facts 1-2, it suffices to prove that $$1 + 2^{-1}\left(\frac{30b+52}{15b-2}\right)^b \ge \left(1 + 2^{-1}\frac{30b+52}{15b-2}\right)^b.$$

We split into two cases:

1. $$b \ge 4$$: It suffices to prove that $$2^{-1}\left(\frac{30b+52}{15b-2}\right)^b \ge \left(1 + 2^{-1}\frac{30b+52}{15b-2}\right)^b$$ or $$2^{-1/b} \ge \frac{15b-2}{30b+52} + 2^{-1}.$$ It is easy and thus omitted (Hint: Take logarithm and then take derivative).

2. $$2 < b < 4$$: Denote $$Q = \frac{30b+52}{15b-2}$$. By the power mean inequality, we have \begin{align} \left(1 + 2^{-1}Q^b\right)^{1/b} &= 9^{1/b}\left(\frac{1^b + (2^{-\frac{4}{b}}Q)^b\cdot 8}{9}\right)^{1/b}\\ &\ge 9^{1/b}\left(\frac{1^2 + (2^{-\frac{4}{b}}Q)^2\cdot 8}{9}\right)^{1/2}. \end{align} It suffices to prove that $$9^{1/b}\left(\frac{1^2 + (2^{-\frac{4}{b}}Q)^2\cdot 8}{9}\right)^{1/2} \ge 1 + 2^{-1}Q$$ or $$9^{2/b}\left(\frac{1^2 + (2^{-\frac{4}{b}}Q)^2\cdot 8}{9}\right) \ge (1 + 2^{-1}Q)^2$$ or $$\frac{1}{9} 9^{2/b} + \frac{8}{9}\left(\frac{9}{16}\right)^{2/b}Q^2 -1 - Q - \frac{1}{4}Q^2 \ge 0.$$ By Facts 3-4, it suffices to prove that $$\frac{1}{9} \cdot \frac{(29b+50)^2}{36(5b-4)^2} + \frac{8}{9}\cdot \frac{363b - 168}{360b + 272} \cdot Q^2 -1 - Q - \frac{1}{4}Q^2 \ge 0$$ or $$\frac{(b-2)(-25504875b^4 + 147431700b^3 - 60455760b^2 - 298781264b + 181422064)}{324(5b-4)^2(45b+34)(15b-2)^2} \ge 0$$ which is true.

We are done.