Inequality proof concerning two variables $k,q$

Question

Let $q,k$ positive integers satisfying the following inequalities $$k\geq3$$ $$q\geq2$$ $$k\geq q$$ I have checked that the following inequality holds for a few values of $k,q$ I have tried but I have not been able to prove it. $$A\triangleq kq{{q^{k-2}}\choose{q^{k-3}+1}}(k-1)\geq kq\cdot q^{2k-4}(q-1)-q^{k-1}(q-1)\frac{(q-1)^3+1}{q}k\triangleq B$$ I tried to simplify $B$ as \begin{eqnarray*} B&=&kq\cdot q^{2k-4}(q-1)-(q^k-q^{k-1})\frac{q^3-3q^2+3q}{q}k\\ &=&kq\cdot q^{2k-4}(q-1)-(q^k-q^{k-1})(q^2-3q+3)k\\ &=&kq\cdot q^{2k-4}(q-1)-(q^{k+2}-3q^{k+1}+3q^k-q^{k+1}+3q^k-3q^{k-1})k\\ &=&kq\cdot q^{2k-4}(q-1)-kq^{k+2}+4kq^{k+1}-6kq^k+3kq^{k-1}\\ &=&kq\cdot(q^{2k-3}-q^{2k-4}-q^{k+1}+4q^k-6q^{k-1}+3q^{k-2})\\ \end{eqnarray*} Now, I need to show that $${{q^{k-2}}\choose{q^{k-3}+1}}(k-1)\geq(q^{2k-3}-q^{2k-4}-q^{k+1}+4q^k-6q^{k-1}+3q^{k-2})$$ and I am stuck at this point.

Answer 1

The inequality does hold for $k\ge 3$, $q\ge 2$ and $k\ge q$.

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We can directly see that the inequality holds for $(k,q)=(3,2),(3,3),(4,2),(4,3),(4,4),(5,2),(6,2)$ respectively.

Here, we use two lemmas.

Lemma 1 : ${{q^{k-2}}\choose{q^{k-3}+1}}(k-1)\ge q^{q^{k-3}}(q-1)$.

Proof for lemma 1 :
$$\begin{align}&{{q^{k-2}}\choose{q^{k-3}+1}}(k-1) \\\\&\ge {{q^{k-2}}\choose{q^{k-3}+1}}(q-1) \\\\&=\frac{q^{k-2}\times (q^{k-2}-1)\times \cdots\times (q^{k-2}-q^{k-3})}{(q^{k-3}+1)\times q^{k-3}\times \cdots \times 1}\times(q-1) \\\\&=\frac{q^{k-2}}{q^{k-3}+1}\times \frac{q^{k-2}-1}{q^{k-3}}\times \frac{q^{k-2}-2}{q^{k-3}-1}\times\frac{q^{k-2}-3}{q^{k-3}-2}\times \cdots\times\frac{q^{k-2}-q^{k-3}}{1}\times (q-1) \\\\&\ge\frac{q^{k-2}}{q^{k-3}+1}\times \frac{q^{k-2}-1}{q^{k-3}}\times q\times q\times\cdots\times q\times q\times (q-1) \\\\&=\frac{q^{k-2}}{q^{k-3}}\times\frac{q^{k-2}-1}{q^{k-3}+1}\times q^{q^{k-3}-1}\times (q-1) \\\\&\ge q\times 1\times q^{q^{k-3}-1}\times (q-1) \\\\&=q^{q^{k-3}}(q-1)\quad\square \end{align}$$

Lemma 2 : $8\cdot q^{2k-3}\geq q^{2k-3}-q^{2k-4}-q^{k+1}+4q^k-6q^{k-1}+3q^{k-2}$.

Proof for lemma 2 : $$\begin{align}8\cdot q^{2k-3}&=q^{2k-3}+4q^{2k-3}+3q^{2k-3} \\\\&\ge q^{2k-3}+4q^{k}+3q^{k-2} \\\\&\ge q^{2k-3}-q^{2k-4}-q^{k+1}+4q^k-6q^{k-1}+3q^{k-2}\quad\square \end{align}$$

Now, from the two lemmas, in order to prove the inequality, it is sufficient to prove $$q^{q^{k-3}}(q-1)\ge 8\cdot q^{2k-3}\tag1$$

We can directly see that $(1)$ holds for $(k,q)=(5,3),(5,4),(5,5),(6,3),(6,4),(6,5),(6,6)$ respectively.

Lemma 3 : $(k-3)\log 2-\log(2k)\ge 0$ for $k\ge 7$.

Proof for lemma 3 : Let $f(k)=(k-3)\log 2-\log(2k)$. Then, $f'(k)=\log 2-\frac 1k$. So, $f(k)$ is increasing for $k\ge 7$ with $f(7)=\log\frac 87\gt 0$. $\ \ \square$

So, for $k\ge 7$, $q\ge 2$ and $k\ge q$, we can get, from the lemma 3, $$\begin{align}\log q\ge \log 2\ge\frac{\log(2k)}{k-3} &\implies(k-3)\log q\ge\log(2k) \\\\&\implies q^{k-3}\ge 2k \\\\&\implies q^{k-3}-2k+3\ge 3 \\\\&\implies q^{q^{k-3}-2k+3}\ge q^3\ge 8 \\\\&\implies q^{q^{k-3}-2k+3}(q-1)\ge 8(q-1)\ge 8 \\\\&\implies q^{q^{k-3}}(q-1)\ge 8\cdot q^{2k-3}\end{align}$$ which is $(1)$.

Therefore, the inequality holds for $k\ge 3$, $q\ge 2$ and $k\ge q$.