If $q$ is prime, then does the following inequality have any solutions for an integer $k > 1$?

Question

Good day!

If $q$ is prime, then does the following inequality have any solutions for an integer $k > 1$?

$$\frac{q + 3}{q} \leq \frac{q^{k+1} - 1}{q^k(q - 1)}$$

Here is my attempt:

$${q^k}(q^2 + 2q - 3) = {q^k}(q - 1)(q + 3) \leq q(q^{k+1} - 1) = q^{k+2} - q$$

$$\Longrightarrow q^{k+2} + 2q^{k + 1} - 3{q^k} \leq q^{k+2} - q \Longrightarrow q(2q^k - 3q^{k-1} + 1) \leq 0 \Longrightarrow q\left(q^{k-1}(2q - 3) + 1\right) \leq 0.$$

The last inequality contradicts $q$ prime with integer $k > 1$.

Is this solution/proof correct? Thanks!

Answer 1

Another way to see that there is no solution to the above inequality is:

$\dfrac{q^{k+1}-1}{q^k(q-1)}< \dfrac{q^{k+1}}{q^k(q-1)}=\dfrac{q}{q-1} = 1+\dfrac{1}{q-1} = 1 + \dfrac{3}{3q-3}= 1 + \dfrac{3}{q+(2q-3)}< 1+\dfrac{3}{q}= \dfrac{q+3}{q}$

Answer 2

Perhaps even easier: starting with $$\frac{q + 3}{q} \leq \frac{q^{k+1} - 1}{q^{k+1} - q^k}\ ,$$ subtract $1$ from both sides, $$\frac3q\le\frac{q^k-1}{q^{k+1}-q^k}\ .$$ Multiplying out denominators, $$3q^k-3q^{k-1}\le q^k-1$$ and so $$1\le q^{k-1}(3-2q)\ .$$ Clearly this is impossible whenever $q\ge\frac32$. It is unimportant that $q$ is prime (or even that $q$ is an integer).