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I need your expertise in solving the following inequality:

Given $n \in \mathbb{Z}^+$, $a \in [0,\infty)$ and $b \in \mathbb{R}^+$ such that $ n > b \geq a$ ($b$ cant be zero).

Can the following be proven:

$$ \frac{1+a}{n - b} \leq \log(n)^c \cdot \left( \frac{1}{n} + \frac{a}{b} \right), $$ where $c \in \mathcal{O}(1)$ (i.e. c can be thought as a constant).

Please advise and thanks in advance.

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    $\begingroup$ The numebr $c$ should be independed on any of $n$, $a$, $b$? And $\log (n)^c$ means $\log (n^c)$ or $(\log n)^c$? $\endgroup$ – Alex Ravsky May 23 '17 at 17:16
  • $\begingroup$ yes and $\log(n)^c = \left( \log(n) \right)^c$. $\endgroup$ – user3492773 May 24 '17 at 18:06
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The answer is negative. Indeed, assume that such $c$ exists. Since both $\frac {1+a}{n-b}$ and $\frac 1n+\frac ab$ are positive, $d=(\log n)^c$ should be positive. We show that $d$ should be zero, obtaining a contadiction. We have

$$d\le \left(\frac 1n+\frac ab\right)\frac {n-b}{1+a}=\frac {na}{b(1+a)}+\frac{1-a}{1+a}-\frac{b}{n(1+a)},$$

and this inequality holds for any admissible $b$. Since when $b$ is growing to $n$ both $\frac {na}{b(1+a)}$ and $-\frac{b}{n(1+a)}$ are diminishing, the inequality holds for each $b$ iff it holds for $b=n$ that is iff

$$d\le \frac {ba}{b(1+a)}+\frac{1-a}{1+a}-\frac{b}{b(1+a)}=0.$$

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