Generalization of a problem, involving radicals and the floor function, proposed by Ramanujan to the Journal of the Indian Mathematical Society The section Solved problems from the Wikipedia Floor and ceiling functions shows several problems proposed by Ramanujan ([1]). The purpose of this post, if possible, is try to get the generalization of some of these identities, for positive integers $n\geq 1$, involving fractions or radicals and the floor function $\lfloor x\rfloor$.
I tried to get a generalization of the identity $(iii)$. I don't know if my conjectural identiy is in the literature or has a good mathematical content, these are my previous failed attempts.
Counterexamples for different formulas.

*

*A counterexample for the (false) identity
$$\lfloor \sqrt[k]{n}+\sqrt[k]{n+1}\rfloor=\lfloor \sqrt[k]{2^k n+k}\rfloor$$
is the integer $n=525$ for the case $k=5$.


*Counterexamples for the (false) identity
$$\lfloor \sqrt[k]{n}+\sqrt[k]{n+1}\rfloor= \left\lfloor \sqrt[k]{2^k n+2(k-1)} \right\rfloor$$
are the integers $n=11$ or $n=610$ for the case $k=6$.
From this thread of experiments I get the following conjecture.
Conjecture. For each integer $k\geq 2$ one has that the identity
$$\lfloor \sqrt[k]{n}+\sqrt[k]{n+1}\rfloor=\left\lfloor 2\sqrt[k]{n+\frac{1}{2}}\right\rfloor$$
holds over integers $n\geq 1$.
I don't know if it is easy to prove, or if you can to find a counterexample.

Question. Do you know if generalizations (thus with a good mathematical meaning, with mathematical significance) of the mentioned problems proposed by Ramanujan are in the literature? In this case, please refer the literature and I try to find and read from the literature these generalizations of $(i)$, $(ii)$ or $(iii)$. In case that aren't in the literature please add yourself generalization, if possible, with its respective proof for some of those identities. In particular, if you know that my conjeture can be proved or can be refuted by finding a counterexample. Many thanks.

Please if some professor/user finds a counterexample it is welcome that he/she comment it, many thanks. I add for example the following scripts written in Pari/GP as a proof of concept/toy model of my conjecture
for(k=2,10,for(n=1, 1000,if(floor((n)^(1/k)+(n+1)^(1/k))!=floor(2*((n+1/2)^(1/k))),print(k," ",n))))
that you can to evaluate from the web Sage Cell Server, just choose GP as language. Thus aren't showed conunterexamples as outputs. Also you've this one
for(k=2,10,for(n=1, 100,print(floor((n)^(1/k)+(n+1)^(1/k)))))
or this
for(k=2,10,for(n=1, 100,print(floor((n)^(1/k)+(n+1)^(1/k))-floor(2*((n+1/2)^(1/k))))))
I add as reference the PARI/GP Developers group of Université Bordeaux 1.
References:
I believe that the corresponding reference is
[1] Srinivasa Ramanujan, Collected Papers, Question 723 in p. 332, Providence RI: AMS / Chelsea (2000).
[2] I've used also the PARI-GP Reference Card (version 2.2.5), by Karim Belabas (2003), based on an earlier version by Joseph H. Silverman.
 A: I can provide a partial proof of your conjecture. It shows that the statement is true for sufficiently large $n$, which in this case means that it is true for $n\geq A_k$, where $A_k$ is a number depending on $k$. Exact form of $A_k$ will be provided in the proof.
We'll need following ingredients:

*

*$\left\lfloor\sqrt[k]{2^kn+2^{k-1}-1}\right\rfloor =\left\lfloor\sqrt[k]{2^kn+2^{k-1}}\right\rfloor$


*$\sqrt[k]{2^kn+2^{k-1}-1}<\sqrt[k]{n}+\sqrt[k]{n+1}$, for $n\geq \left\lceil A_k\right\rceil$, where $\left\lceil A_k\right\rceil=\left\{\begin{array}{ll}
0 & \text{for }k=1 \\
2^{k-3} & \text{for }k\geq2
\end{array}\right.$


*$\sqrt[k]{n}+\sqrt[k]{n+1}<\sqrt[k]{2^kn+2^{k-1}}$
These three ingredients imply our conjecture. Proofs of individual ingredients provided below.

*

*Note that ${2^kn+2^{k-1}-1}$ and ${2^kn+2^{k-1}}$ differ only by 1, so the only
possibility for 1. not being true is when $\sqrt[k]{2^kn+2^{k-1}}$ is an integer. If we
assume that
$$\sqrt[k]{2^kn+2^{k-1}}=q,\quad\text{for }q\in\mathbb{N},$$
then
$$2^kn+2^{k-1}=2^{k-1}(2n+1)=q^k.$$
We can notice, that $q^k$ is divisible by $2^{k-1}$, but this implies that $2^k|q^k$,
which can't be true since $2n+1$ is odd. Hence 1. is true.


*Let's rewrite 2. a little bit. We want to prove that
$$\frac{\sqrt[k]{n}+\sqrt[k]{n+1}}{2}>\sqrt[k]{n+\frac{2^{k-1}-1}{2^k}}.$$
By applying AM-GM inequality we obtain
$$\frac{\sqrt[k]{n}+\sqrt[k]{n+1}}{2}>\sqrt[2k]{n^2+n}.$$
The inequality
$$\sqrt[2k]{n^2+n}\geq\sqrt[k]{n+\frac{2^{k-1}-1}{2^k}}$$
occurs if and only if $n\geq2n\dfrac{2^{k-1}-1}{2^k}+\left(\dfrac{2^{k-1}-1} 
{2^k}\right)^2$, but this inequality implies that
$$n\geq \dfrac{2^{2(k-1)}+1-2^k}{2^{k+1}}=:A_k,$$
so if $n\geq \left\lceil A_k\right\rceil$, then 2. is true.


*Let's rewrite 3. in the same manner as we've rewritten 2. We want to prove that
$$\frac{\sqrt[k]{n}+\sqrt[k]{n+1}}{2}<\sqrt[k]{n+\frac{1}{2}}.$$
By Generalized mean inequality for $k$ exponent we obtain
$$\frac{\sqrt[k]{n}+\sqrt[k]{n+1}}{2}<\sqrt[k]{\frac{n + n + 1}{2}}=\sqrt[k] 
{n+\frac{1}{2}}.$$
The equality can't occur, because $\sqrt[k]{n}\neq \sqrt[k]{n+1}$ for all
$n\in\mathbb{N}$. $$\tag*{$\blacksquare$}$$
The cases, when $n<A_k$ of course can be checked by computer, but that's not a full proof. From now on I'm working on finding the full proof of it.
