TITLE: Inequality for functions on [0,1]
QUESTION [10 upvotes]: Let $a\in (0,1), \;\;\psi_a(x):=\prod_{j=0}^\infty (1-a^{2j+1}x).$
Question. Is it true that, for all $x\in [0,1]$ and all $k\in\mathbb{N},$ the following inequality holds:
$$\frac{x^k}{(1-a)(1-a^3)\dots (1-a^{2k-1})}\leq\frac{1}{\psi_a(x)}\quad ?$$
For $a\leq 0.5,$ this is obvious. When $a$ approaches 1, it becomes difficult (for me).
The inequality arises in the study of the dependence of the limit $q$-Bernstein operator on parameter $q.$

REPLY [2 votes]: This was meant to cover the "calculus part" in fedja's initial answer (now completed independently), $0\le a\le a_0:=(1/9+1/25)^{1/3}$, and may possibly be completed to an independent proof.
The functions $\phi(x):=\prod_{j=0}^\infty {1\over 1-a^{2j+1}x}$ and $x^{-k}$ are positive and logarithmically convex on $[0,1]$, therefore such is their product $x^{-k}\phi(x)$ too, which is therefore convex. Moreover $\big(x^{-k}\phi(x)\big)'_{x=1}=-k\phi(1)+\phi'(1)\le0$ iff $k\ge \phi'(1)/\phi(1)=\sum_{j=0}^\infty{a^{2j+1}\over 1-a^{2j+1}},$ in which case $x^{-k}\phi(x)$ has a minimum at $x=1$. We conclude that the improved inequality
$$\prod_{j=0}^\infty {1\over 1-a^{2j+1}x}\ge \Big( \prod_{j=0}^\infty {1\over 1-a^{2j+1}}\Big)x^k,\qquad 0\le x\le 1$$
holds whenever $k\ge \sum_{j=0}^\infty{a^{2j+1}\over 1-a^{2j+1}};$ in particular, (by a simple numeric estimates of the sum) for all $k\ge 2$ and $0\le a\le a_0$. (For $k=1$ the inequality can be proven directly as shown in  T. Amdeberhan's answer)