During a math class, a proposed problem on a handout is as follows:

Prove that \[\lfloor x\rfloor +\dfrac{\lfloor 2x\rfloor}{2}+\dfrac{\lfloor 3x\rfloor}{3}+\cdots +\dfrac{\lfloor nx\rfloor}{n}\le \lfloor nx\rfloor\] for all positive reals \(x\) and integer \(n\).

Can someone help me prove it? I tried to use a periodic argument, proving it true for \(x\in [0,1)\) then plugging in \(x=x'+k\) for positive integer \(k\) to prove the inequality for all other \(x\). However it didn't work, because my inequality substitutions were too strict and I ended up with \((n-1)x\le \lfloor nx\rfloor\) which is not true.

Thanks!

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TopNewesthttp://mks.mff.cuni.cz/kalva/usa/usoln/usol815.html – Mathh Mathh · 2 years, 11 months ago

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I don't know if I am right or wrong ... So please correct me ... Thanks in advance ... here is my Solution : As we know, floor(x) = x - {x} , where {x} is fractional part. if we replace all the floor values in the above eqn will reduce to

LHS : nx - [ {x} + ({2x}/2) + ({3x}/3) + ({4x}/4) + ........... + ({nx}/n) ]

What is interesting to see is that {x} = {2x}/2 = {3x}/3 = ..... = {nx}/n (You can check it with any value of x :) )

Hence LHS will become : nx - n{x}

Solving RHS we got : floor(nx) = nx - {nx} As the fractional part of {nx} will always lie between 0 and 1, i.e. 0 <= {nx} < 1

And it is easy to see that the product of n and {x} in LHS is always greater than or equal to (in case of 0 as fractional value) {nx} in RHS.

Probably this will prove it ... What say guys ??? – Vipin Kumar · 2 years, 11 months ago

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– Abhinav Raichur · 2 years, 11 months ago

that should prove it man :)...... i don't find any serious anomaly..... good jobLog in to reply

– Mathh Mathh · 2 years, 11 months ago

{x} = {2x}/2 = {3x}/3 = ..... = {nx}/n is false (see my other comments).Log in to reply

– Abhinav Raichur · 2 years, 11 months ago

yes and it is just the special case where inequality turns to equality...... it gives maximum value of LHSLog in to reply

– Mathh Mathh · 2 years, 11 months ago

\(n\{x\}\neq \{nx\}\). Try, e.g., \(x=0.6\), \(n=2\). Then \(2\cdot \{0.6\}=1.2\neq\{2\cdot 0.6\}=0.2\).Log in to reply

– Vipin Kumar · 2 years, 11 months ago

When did I say n{x} = {nx} ??? Please check it once more :)Log in to reply

I could have equally said that \(\{x\}\neq \frac{\{nx\}}{n}\), just multiply both sides of your equality by \(n\) (we can, since \(n\neq 0\)). Both of the equalities are equivalent. – Mathh Mathh · 2 years, 11 months ago

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– Vipin Kumar · 2 years, 11 months ago

Yups got my mistake... thank you... :)Log in to reply

@Daniel Liu @Finn Hulse : Guys can you please check the above solution ??? – Vipin Kumar · 2 years, 11 months ago

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@Finn Hulse

I still don't think it proves it.Your solution depends on the fact that \[\{x\}+\dfrac{\{2x\}}{2}+\cdots+\dfrac{\{nx\}}{n}\ge \{nx\}\] but you didn't prove that. – Daniel Liu · 2 years, 11 months ago

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– Finn Hulse · 2 years, 11 months ago

Meh I'm tired. :PLog in to reply

– Finn Hulse · 2 years, 11 months ago

It's good as far as I can tell! :DLog in to reply

I think using induction approach should also work – Roger Lu · 2 years, 11 months ago

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Apparently this is USAMO 1981 #5. – David Lee · 2 years, 11 months ago

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– Daniel Liu · 2 years, 11 months ago

Yep.Log in to reply

– Kaan Dokmeci · 2 years, 11 months ago

I was just gonna say...Log in to reply