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I think it's hard but let me propose it :

Let $x\geq 0$ then we have : $$\ln(\ln(\ln(\ln(\ln(\ln(\ln(\ln(x+1)+1)+1)+1)+1)+1)+1)+1)\leq \operatorname{W}(\operatorname{W}(\operatorname{W}(\operatorname{W}(x))))$$

Where we speak about the Lambert's function .

To prove it I have tried to denest the logarithm since the inverse function of $f(x)=\ln(x+1)$ is $f^{-1}(x)=e^x-1$ it gives a nice power tower . Then I use the fact that :

$$\forall x>0\quad e^{\operatorname{W}(x)}=\frac{x}{\operatorname{W}(x)}$$

Wich is just the definition of the Lambert's function .

Unfortunately we can use it just one time and I got stuck here .

We can do the inverse and try to work with $p(x)=xe^x$ to eliminate the special function and keep just elementary function . Again we got a power tower but it's delicate to derivate it .

If you have a hint or a proof ...

...Thanks in advance

Simply Beautiful Art
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Barackouda
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For $t \geqslant 1,$ because the function $t \mapsto 1/t$ is convex, $$ \ln t \leqslant \frac{t - 1}2\left(1 + \frac1t\right) = \frac{t^2 - 1}{2t}, $$ therefore $$ t\ln t \leqslant \frac{t^2 - 1}2 = (t - 1) + \frac{(t - 1)^2}2 \leqslant e^{t - 1} - 1 \quad (t \geqslant 1). $$ Equivalently, because the function $t \mapsto e^{t - 1} - 1$ is strictly increasing on $[1, \infty),$ $$ (\ln(x + 1) + 1)\ln(\ln(x + 1) + 1) \leqslant x \quad (x \geqslant 0). $$ Therefore $\ldots$

Calum Gilhooley
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