On q-analogs of integer inequalities

Question

I've been studying properties of q-analogs recently and had some difficulties on trying to prove some basic-looking inequalities. For $q \in (0, 1)$ and $n\in\mathbb{N}$ let's define q-integer $[n]_q = \frac{1-q^n}{1-q}$. I understand that it is easy to prove that, for example, $[n]_q < [n+1]_q$ (as it is equivalent to $q^{n+1} < q^n$). But what about quadratic inequalities? As a certain example, let $k \in \mathbb{N}$ s.t. $k+1 \leq n$. For "ordinary" integers, it is easy to show that $ \frac{k(k+1)}{(n-k)(n-k-1)} \leq 1 $ for $2k \leq n-1$ since $$ \frac{k(k+1)}{(n-k)(n-k-1)} \leq 1 \Leftrightarrow k^2 + k \leq n^2 - 2 n k + k^2 - n + k \Leftrightarrow n^2 - 2n k - n \geq 0 \Leftrightarrow \\ \Leftrightarrow n - 2k - 1 \geq 0 \Leftrightarrow n \geq 2k + 1 $$ However, if we change all integers in this equality to q-integers, we will need to prove that $$ \frac{ (1 - q^k) (1 - q^{k+1}) } { (1 - q^{n-k}) (1 - q^{n-k-1}) } \leq 1 $$ and I can't come up with idea to prove it, since $[k]_q [k+1]_q = [k^2 + k]_q$ is false. However, my gut feeling and numerical checks show that these must indeed hold (and for other similar inequalities also). As an argument (weak, as I admit), it is known that q-integers are, in fact, polynomials in $q$ and $\lim_{q \to 1-} [n]_q = n$, so due to "niceness" of polynomial functions it would be rather strange if some inequality held only in limit.

Answer 1

One has $$\begin{align}&\frac{(1 - q^k) (1 - q^{k+1})}{(1 - q^{n-k}) (1 - q^{n-k-1})} \leqslant 1 \\\\&\iff (1 - q^k) (1 - q^{k+1})\leqslant (1 - q^{n-k}) (1 - q^{n-k-1}) \\\\&\iff 1-q^{k+1}-q^k+q^{2k+1}\leqslant 1-q^{n-k-1}-q^{n-k}+q^{2n-2k-1} \\\\&\iff q^{k+1}+q^k-q^{2k+1}-q^{n-k-1}-q^{n-k}+q^{2n-2k-1}\geqslant 0\end{align}$$

Now, let $$f(k):=q^{k+1}+q^k-q^{2k+1}-q^{n-k-1}-q^{n-k}+q^{2n-2k-1}$$

Then, one has $$f(k)\gt f(k+1)\tag1$$ since $$\begin{align}f(k)-f(k+1)&=q^{k+1}+q^k-q^{2k+1}-q^{n-k-1}-q^{n-k}+q^{2n-2k-1} \\&\qquad -\bigg(q^{k+2}+q^{k+1}-q^{2k+3}-q^{n-k-2}-q^{n-k-1}+q^{2n-2k-3}\bigg) \\\\&=q^k(1-q^{2})-q^{2k+1}(1-q^{2})+q^{n-k-2}(1-q^{2})-q^{2n-2k-3}(1-q^{2}) \\\\&=(1-q^2)\bigg(q^k-q^{2k+1}+q^{n-k-2}-q^{2n-2k-3}\bigg) \\\\&=(1-q^2)\bigg(q^k(1-q^{k+1})+q^{n-k-2}(1-q^{n-k-1})\bigg) \\\\&\gt 0\end{align}$$

Also, one has $$f\bigg(\left\lfloor\frac{n-1}{2}\right\rfloor\bigg)\geqslant 0\tag2$$ since for $n=2m$, $$f(m-1)=q^{m-1}(1-q^2)(1-q^{m})\geqslant 0$$ and for $n=2m+1$, $$f(m)=q^{m+1}+q^m-q^{2m+1}-q^{m}-q^{m+1}+q^{2m+1}=0$$

Therefore, if $k\leqslant\dfrac{n-1}{2}$, then it follows from $(1)(2)$ that $$f(k)\geqslant 0.$$