today I've encountered a question which was like the following;

$\dfrac{-1}{3}<a<\dfrac{1}{5}$ and $\dfrac{1}{6}<b<\dfrac{1}{5}$, what is the minimum integer value $\dfrac{a+b}{ab}$ can get?

First of all I've noticed that $\dfrac{a+b}{ab}=\dfrac1a+\dfrac1b$....(1)

Then I've tried to turn the inequalities to $\dfrac1a$ and $\dfrac1b$, doing it for $b$ was easy, I quickly got $6>b>5$, however when I tried to do it for $a$ it was problematic. Due to the fact that there is a negative number in the LHS. I've tried to separate $a$ in parts like $-3<\dfrac1a\leq -1$ but I then realised that this would not be a legal move...

What are your suggestions?


$$E:={a+b\over ab} ={1\over a}+{1\over b} $$

While $a\to 0 $ and $a<0$ and $b=fixed$ we have ${1\over a} \to -\infty$ so $E\to -\infty$. So $E$ does not have minimum.

  • $\begingroup$ Thank you, but I don't see where you used the inequalities about $a$ and $b$. $\endgroup$ – Deniz Tuna Yalçın Dec 1 '17 at 16:18
  • $\begingroup$ They are of no use since $a$ is ''moving'' around zero. So ${1\over a}$ can be arbitrary big if $a$ is positive and goes to 0 and arbitrary small if $a$ is negative and goes to 0 $\endgroup$ – Aqua Dec 1 '17 at 16:19
  • $\begingroup$ Draw a graph $f(x)={1\over x}$. $\endgroup$ – Aqua Dec 1 '17 at 16:21
  • $\begingroup$ I see now, so the question is flawed. The same result would be got if there were sth like; $\dfrac{1}{3}<a<\dfrac{1}{5}$ is that also correct? Thank you by the way:) $\endgroup$ – Deniz Tuna Yalçın Dec 1 '17 at 16:22
  • $\begingroup$ No, in that case E > 5+5 =10 $\endgroup$ – Aqua Dec 1 '17 at 16:22

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