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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?

2 Answers 2

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we have $$(2n-1)a+b

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    What if $(1-an^2)$ is negative? Then the product and therefore one of the two summands is negative. – 2017-01-07
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    but $$1-b$$ is also negative! – 2017-01-07
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    No, since b<1 we have 0<1-b. – 2017-01-07
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    one moment i'm thinking – 2017-01-07
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    yea that is cleat – 2017-01-07
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    now it works i will post my now proof – 2017-01-07
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    Now there's a $-b^2$ missing in the numerator of the last fraction, which again makes in more negative – 2017-01-07
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    i will have a look at this – 2017-01-07
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    no i think it is all ok, which step do you mean? – 2017-01-07
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    You just fixed it ;) – 2017-01-07
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    thank you for your discussion! – 2017-01-07
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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 $n$).

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    I like the idea of using the discriminant to attempt this problem. – 2017-01-07