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Let $a,b,c$ and $d$ be positive real numbers such that $cd=1$ Prove that there is an integer,$n$, such that $ab < n^2<(a+c)(b+d)$

I'm having trouble solving this question, I would like to get a hint to help me start off this question. I attempted it by trying to expand and substitute but that's not helping me find n and i'm pretty clueless on what to do next. I normally have trouble starting off my questions... I know this isn't an advice forum but if possible are there any tips or thought process that I could follow to start figuring out how to start math problems?

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The assertion is false. Take $a=b=c=d=1$. Then

$ab=1$ and $(a+c)(b+d)=4$

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    OP means that for certain conditions imposed on $a,b,c,d$, the conclusion can be made to be true. So a counter example in this case is not the best thing to pursue. – 2017-01-08
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    ????? Math4 life wrote: "Let $a,b,c$ and $d$ be positive real numbers such that $cd$=1 Prove that there is an integer,$n$, such that $ab – 2017-01-08
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    @DeepSea How could you possibly know the OP meant *that*?? Even when he didn't write anything even close to that. – 2017-01-08
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The proof uses the following property of real number: If $x > 0, y > 0, y > x, y - x \ge 1$, then $(x,y)$ contains a natural number $n$.I can prove this for you, but I want to use it to prove your question first. In order for the proof to work, we must assume further that $\sqrt{ab} \notin \mathbb{N}$, and we have: $(a+c)(b+d) = (a+c)(b+\dfrac{1}{c})= ab+\dfrac{a}{c} + bc + 1 \ge (\sqrt{ab})^2+ 2\sqrt{ab} + 1 = (\sqrt{ab} + 1)^2$, by AM-GM inequality. Thus we can choose $n \in (\sqrt{ab}, \sqrt{ab} + 1)\implies ab < n^2 < (a+c)(b+d)$ as claimed.

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    If $a=b=1$ then each $n \in (\sqrt{ab}, \sqrt{ab} + 1)=(1,2)$ is not an integer ! – 2017-01-08
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    I know that but if $\sqrt{ab} \notin \mathbb{N}$ , then it is true. The point here is NOT about counter example. – 2017-01-08