I am trying to prove this inequality for real $a$ and $b$ with $0\lt a \lt b \lt 1$ and integer $n \ge 0$:
$$(2n-1)a+b \lt n^2ab+1$$
I tried using induction but that approach failed. Also I tried making the smaller side greater and proving this and making the bigger side smaller and prove that, but neither of them worked. Any ideas?
On
The inequality can be rewritten as: $\;ab \,n^2 -2a\,n +a-b+1 \gt 0$.
The quadratic in $n$ has the dominant coefficient positive $ab \gt 0\,$, and its discriminant is negative given that $0 \lt a \lt b \lt 1$:
$$ \frac{1}{4}\Delta = a^2 - ab(a-b+1) = a^2-a^2b-ab(1-b) = a(a-b)(1-b) \;\;\lt\;\; 0 $$
Therefore the inequality holds for all $\forall \,n \in \mathbb{R}\,$ (and in particular for all non-negative integers $n$).
we have $$(2n-1)a+b<n^2ab+1$$ with $$0<a<b<1$$ this is equivalent to $$0<n^2ab-(2n-1)a+1-b$$ and this is equivalent to $$0<n^2-\frac{2n-1}{b}+\frac{1-b}{ab}$$ and this is equivalent to $$0<n^2-\frac{2n}{b}+\frac{1}{b^2}+\frac{1}{b}+\frac{1-b}{ab}-\frac{1}{b^2}$$ further $$0<(n-\frac{1}{b})^2+\frac{(1-b)(b-a)}{ab^2}$$ which is obviously true.