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@Calvin Lin
–
I started by assuming d<b+f. Then I did lot of manipulations and after a while a wrong result came up which means this is a contradiction. So d must be >=b+f. Is this correct?

@Kushagra Sahni
–
Assuming that the steps are correct, then that sounds like a possible proof approach. You should write up the solution and others can comment on it.

It is easy to get bc-ad>0.b,c,a,d,e,f are integers.So I was wondering how can I use it in the problem?Then I thought that bc-ad>=1 might help as after all I have to show something >= something.So using bc-ad>=1 is better than using bc-ad>0.Hence I used it.Now I broke (dbe - bfc) in the following way so as to use the fact
bc-ad>=1 & de-fc>=1 & finally arrived the solution.

@Rajdeep Brahma
–
Wonderful! Thanks for explaining. This helps to demystify what you did. Previously when stated as 1 sentence, it seems like "I magically created this equation and got the answer".

You've now added "Here is my reasoning for why I considered these steps, and how I was wanting to use the conditions of the problem". This will help @Md Zuhair (and everyone else) learn from this and apply it to other scenarios.

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## Comments

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TopNewest$\dfrac{a}{b} < \dfrac{c}{d} < \dfrac{e}{f} ;~ af - be = -1 \\ \dfrac{a}{b} < \dfrac{c}{d} \\ ad < bc \\ \boxed{ad + x = bc} \\ \dfrac{c}{d} < \dfrac{e}{f} \\ cf < de \\ \boxed{cf + y = de\\ cf = de - y} \\ \boxed{af - be = -1 \\ af = be-1}\\ ad + x = bc \\ adf + xf = bcf \\ d(af) + xf = b(cf) \\ d(be-1) + xf = b(de-y) \\ dbe - d + xf = dbe - by \\ xf - d = -by \\ d = xf + by \\ \boxed{d \ge b + f}$

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You are going to top TN Region RMO This time @Vicky Vignesh

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Great! The interesting fact is that $d = xf + by$, which is extremely restrictive.

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Yeah, restricts $x, y \in R$

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What have you tried? What do you know?

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I did'nt understood your question?

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How have you tried to solve this problem?

What observations have you made?

For example, do you know if $d > b$ or $d > f$?

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@Vicky Vignesh .. Could you?

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If you've tried nothing, then go try something. What approaches can you think of? What comparisons can we make?

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Note that bc-ad>=0.We have bc-ad>=1.Similarly de-fc>=1.So d=d(be-af)=dbe-daf=dbe-bfc+bfc-adf=b(de-fc)+f(bc-ad)>=b+f This might be a shorter proof.

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Very nicely done. What is the motivation / intuition behind how you arrived at this solution? What led you to think about it?

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It is easy to get bc-ad>0.b,c,a,d,e,f are integers.So I was wondering how can I use it in the problem?Then I thought that bc-ad>=1 might help as after all I have to show something >= something.So using bc-ad>=1 is better than using bc-ad>0.Hence I used it.Now I broke (dbe - bfc) in the following way so as to use the fact bc-ad>=1 & de-fc>=1 & finally arrived the solution.

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You've now added "Here is my reasoning for why I considered these steps, and how I was wanting to use the conditions of the problem". This will help @Md Zuhair (and everyone else) learn from this and apply it to other scenarios.

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Well, I couldnt do this :P

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check out the third problem-http://www.isibang.ac.in/~statmath/olympiad/3sol.pdf

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