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?


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.

  • $\begingroup$ What if $(1-an^2)$ is negative? Then the product and therefore one of the two summands is negative. $\endgroup$ – MinecraftShamrock Jan 7 '17 at 14:41
  • $\begingroup$ but $$1-b$$ is also negative! $\endgroup$ – Dr. Sonnhard Graubner Jan 7 '17 at 14:42
  • $\begingroup$ No, since b<1 we have 0<1-b. $\endgroup$ – MinecraftShamrock Jan 7 '17 at 14:44
  • $\begingroup$ one moment i'm thinking $\endgroup$ – Dr. Sonnhard Graubner Jan 7 '17 at 14:44
  • $\begingroup$ yea that is cleat $\endgroup$ – Dr. Sonnhard Graubner Jan 7 '17 at 14:45

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 $n$).

  • 1
    $\begingroup$ I like the idea of using the discriminant to attempt this problem. $\endgroup$ – MinecraftShamrock Jan 7 '17 at 23:40

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