Is it possible to get a tight lower bound for $k$, if the following inequality holds?

Question

Is it possible to get a tight lower bound for $k$ (in terms of $q$), if the following inequality holds? $$\frac{q^2 - 1}{q^2} < \frac{q^{k+1}}{q^{k+1} + 1}$$ Here, $q$ is a prime number, and $k$ is a positive integer.

This question arises out of considerations surrounding this answer to a closely related MSE question.

Here are some data on the corresponding lower bounds for $k$ obtained from this inequality on various primes $q$:

$$ \begin{array}{|c|c|c|} \hline q& \text{Lower bound for } k & \text{Approximate value of lower bound} \\ \hline 2& \log_2(3/2)& 0.584963 \\ \hline 3& \log_3(8/3)& 0.892789 \\ \hline 5& \log_5(24/5)& 0.974636 \\ \hline 7& \log_7(48/7)& 0.989404 \\ \hline 11& \log_{11}(120/11)& 0.996539 \\ \hline 13& \log_{13}(168/13)& 0.997686 \\ \hline \end{array} $$

(I used WolframAlpha to manually pull these data one by one.)

From the first few values on the lower bounds for $k$ obtained from the inequality by plugging in the values of the first six primes, I hereby conjecture that:

CONJECTURE: If the following inequality holds $$\frac{q^2 - 1}{q^2} < \frac{q^{k+1}}{q^{k+1} + 1},$$ where $q$ is a prime number and $k$ is a positive integer, then $$k > \log_{q}(q - \frac{1}{q}).$$

Here are my:

QUESTIONS: Does this conjecture hold for all primes $q$? If so, can you prove it?

Answer 1

If $1≠q>0, ~k>0$, then

$$\frac{q^2 - 1}{q^2} < \frac{q^{k+1}}{q^{k+1} + 1}$$

$$\frac{q^{k+1} - q^2 + 1}{q^2 (q^{k+1} + 1)}>0$$

$$q^{k + 1}+ 1>q^2$$

$$k+1>\log_q(q^2-1)$$

$$k>\log_q(q^2-1)-1$$

$$k>\log_q(q^2-1)-\log_q q$$

$$k>\log_q\left(\frac{q^2-1}{q}\right)$$

$$k>\log_q\left(q-\frac{1}{q}\right).$$

This means, $q$ need not be prime and $k$ need not be integer.

Answer 2

\begin{align} &\frac{q^2 - 1}{q^2} < \frac{q^{k+1}}{q^{k+1} + 1}\\ \iff&1-\frac{q^2 - 1}{q^2}>1-\frac{q^{k+1}}{q^{k+1} + 1}&&(\because a<b\iff 1-a>1-b)\\ \iff&\frac1{q^2}>\frac1{q^{k+1}+1}\\ \iff&q^{k+1}>q^2-1&&(\because (q^2)(q^{k+1}+1)>0)\\ \iff&k>\log_q(q^2-1)-1&&(\because\text{Taking $\log_q$ both sides, since $q>1$})\\ \iff&k>\log\left(\dfrac{q^2-1}{q}\right)&&\left(\because \log_ab-\log_ac = \log_a\frac bc\right)\\ \iff&k>\log\left(q-\frac1q\right) \end{align}

Answer 3

Since $q$ is a prime number, the inequality under consideration is equivalent to $$q^{k+3} - q^{k+1} + q^2 - 1 < q^{k+3} \iff q^{k+1} > q^2 - 1$$ $$\iff q^k > \frac{q^2 - 1}{q} \iff k > \log_{q}(q - \frac{1}{q}).$$

This shows that the Conjecture in the OP is true for all primes $q$.