If \(a,b,c,d,e,f\) are positive integers and \(a/b<c/d<e/f\).

Suppose \(af-be=-1\) ,then prove that \(d>= b+f \).

If \(a,b,c,d,e,f\) are positive integers and \(a/b<c/d<e/f\).

Suppose \(af-be=-1\) ,then prove that \(d>= b+f \).

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TopNewest\(\dfrac{a}{b} < \dfrac{c}{d} < \dfrac{e}{f} ;~ af - be = -1 \\ \dfrac{a}{b} < \dfrac{c}{d} \\ ad < bc \\ \boxed{ad + x = bc} \\ \dfrac{c}{d} < \dfrac{e}{f} \\ cf < de \\ \boxed{cf + y = de\\ cf = de - y} \\ \boxed{af - be = -1 \\ af = be-1}\\ ad + x = bc \\ adf + xf = bcf \\ d(af) + xf = b(cf) \\ d(be-1) + xf = b(de-y) \\ dbe - d + xf = dbe - by \\ xf - d = -by \\ d = xf + by \\ \boxed{d \ge b + f} \) – Vicky Vignesh · 4 months ago

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– Calvin Lin Staff · 4 months ago

Great! The interesting fact is that \( d = xf + by \), which is extremely restrictive.Log in to reply

– Vicky Vignesh · 4 months ago

Yeah, restricts \(x, y \in R\)Log in to reply

@Vicky Vignesh – Md Zuhair · 4 months ago

You are going to top TN Region RMO This timeLog in to reply

What have you tried? What do you know? – Calvin Lin Staff · 4 months ago

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– Md Zuhair · 4 months ago

I did'nt understood your question?Log in to reply

What observations have you made?

For example, do you know if \( d > b \) or \( d > f \)? – Calvin Lin Staff · 4 months ago

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– Md Zuhair · 4 months ago

I could'nt do it.Log in to reply

@Vicky Vignesh .. Could you? – Md Zuhair · 4 months ago

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If you've tried nothing, then go try something. What approaches can you think of? What comparisons can we make? – Calvin Lin Staff · 4 months ago

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– Kushagra Sahni · 4 months ago

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?Log in to reply

– Calvin Lin Staff · 4 months ago

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.Log in to reply

– Vicky Vignesh · 4 months ago

You must use the given equality with the inequality. Try adding any variable to rest of inequality with a'equal to' symbolLog in to reply