TITLE: 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 $b>2$.
QUESTION [14 upvotes]: 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.

REPLY [2 votes]: 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:

$b \ge 4$: 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).

$2 < b < 4$: 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.