Let \(x,y\) be positive real numbers. Prove that \[\dfrac{\lfloor x\rfloor+\lfloor y\rfloor}{\lfloor x+y\rfloor}\le \dfrac{\lceil x\rceil+\lceil y\rceil}{\lceil x+y\rceil}\]

Where \(\lfloor \ldots\rfloor\) is the greatest int function and \(\lceil \ldots \rceil\) is the smallest int function.

In general, prove that for reals \(x_1,x_2\ldots x_n\), that \[\dfrac{\sum\limits_{i=1}^{n}\lfloor x_i\rfloor}{\left\lfloor\sum\limits_{i=1}^{n} x_i\right\rfloor}\le \dfrac{\sum\limits_{i=1}^{n}\lceil x_i\rceil}{\left\lceil\sum\limits_{i=1}^{n} x_i\right\rceil}\]

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\[\lfloor{n}\rfloor \leq \lceil{n}\rceil \leq \lfloor{n}\rfloor + 1\] – Sharky Kesa · 2 years, 9 months ago

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– Finn Hulse · 2 years, 9 months ago

I think the \(<\) should be changed to a \(\leq\).Log in to reply

– Sharky Kesa · 2 years, 9 months ago

No, cos it isn't possible.Log in to reply

– Finn Hulse · 2 years, 9 months ago

\(\lceil{3.5}\rceil=\lfloor{3.5}\rfloor+1 \longrightarrow 4=3+1\).Log in to reply

– Sharky Kesa · 2 years, 9 months ago

Sorry, didn't realise.Log in to reply

– Finn Hulse · 2 years, 9 months ago

It's okay bro. I like your spelling of "realize". :DLog in to reply

– Sharky Kesa · 2 years, 9 months ago

It's realise for me. We have to use British English. You have to use your Americanised version of English. Also, when writing by hand, s takes shorter time to write than z in cursive.Log in to reply

– Finn Hulse · 2 years, 9 months ago

Yeah bro, it's okay I know. Wait do you write cursive? :OLog in to reply

– Sharky Kesa · 2 years, 9 months ago

Yes. We have to write in cursive/running writing. It's faster.Log in to reply

Cool problem! I think this is a valid approach:

Looking at the first equation, we see that equality is attained if \(x\) and \(y\) are both integers, which would make the equation just \(1 \leq 1\). Split this into two cases:

Case 1:

\(\lfloor{x}\rfloor+\lfloor{y}\rfloor < \lfloor{x+y}\rfloor\). In this case, obviously neither \(x\) or \(y\) are integers otherwise their sum would be the same as the floor of their sum. Thus the numerator of the RHS will become \(\lfloor{x}\rfloor+1+\lfloor{y}\rfloor+1\) and denominator \(\lfloor{x}\rfloor+\lfloor{y}\rfloor+1\). Obviously the LHS will be \(<1\) and the RHS \(>1\) so the inequality is held. Can we generalize this inequality for the general case? Yes, because for each extra term the LHS will still be \(<1\) and the denominator \(>1\) and the reverse on the RHS.

Case 2:

Both \(x\) and \(y\) are integers. The inequality becomes

\[\dfrac{x+y}{x+y} \leq \dfrac{x+y}{x+y}\]

which is obviously true. You'll see how it's really just an extension of Case 3. It will clearly also be true even for the general case given.

Case 3:

\(\lfloor{x}\rfloor+\lfloor{y}\rfloor = \lfloor{x+y}\rfloor\). This will be true as long as the fractional part of \(x\) and \(y\) do not have a sum \(>1\), unlike Case 1. So, like, \(x\) could be \(3.3\) and \(y\) could be \(1.6\), and \(0.6+0.3 \ngeq 1\). In this case, the inequality becomes

\[\dfrac{x+y-\text{frac}(x)-\text{frac}(y)}{x+y-\text{frac}(x)-\text{frac}(y)} \leq \dfrac{x+y+1-\text{frac}(x)-\text{frac}(y)}{x+y+1-\text{frac}(x)-\text{frac}(y)}\]

where \(\text{frac}(n)\) represents the fractional part, or just \(n-\lfloor{n}\rfloor\). Obviously, both sides are equal and the inequality is satisfied. We see that for all \(x_1, x_2, x_3, \dots\) this is satisfied as well since there will always be that extra \(+1\) in the numerator and denominator of the RHS since the sum of the fractional parts is \(<1\).

It is impossible for \(\lfloor{x}\rfloor\) to be greater than \(\lfloor{x+y}\rfloor\) because the largest possible value of \(\text{frac}(x)+\text{frac}(y)=0.999\dots+0.999\dots\) which is \(<2\).

Quod erat demonstratum. – Finn Hulse · 2 years, 9 months ago

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Let \(x=x_1+x_f\) and \(y=y_1+y_f\) where \(x_1,y_1\) are integers and \(x_f,y_f\) are the fractional part of \(x,y\) respectively.

Note that \[\lfloor x\rfloor +\lfloor y\rfloor \le \lfloor x+y\rfloor \] because \[x_1+y_1 \le x_1+y_1+\lfloor x_f+y_f\rfloor\]

This equation becomes \[\dfrac{\lfloor x\rfloor +\lfloor y\rfloor}{\lfloor x+y\rfloor}\le 1\]

Also, note that \[\lceil x \rceil +\lceil y \rceil \ge \lceil x+y\rceil\] because \[x_1+1+y_1+1\ge x_1+y_1+\lceil x_f+y_f\rceil\]

This equation becomes \[\dfrac{\lceil x \rceil +\lceil y \rceil}{\lceil x+y\rceil}\ge 1\]

Putting the two equations together, we have \[\dfrac{\lfloor x\rfloor +\lfloor y\rfloor}{\lfloor x+y\rfloor}\le 1\le \dfrac{\lceil x \rceil +\lceil y \rceil}{\lceil x+y\rceil}\]

and we are done. \(\Box\) – Daniel Liu · 2 years, 9 months ago

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– Mathh Mathh · 2 years, 9 months ago

How can you claim that \(\lfloor x\rfloor + \lfloor y\rfloor\le \lfloor x+y\rfloor\implies \frac{\lfloor x\rfloor + \lfloor y\rfloor}{\lfloor x+y\rfloor}\le 1\) given that \(\lfloor x+y\rfloor\) can be negative? I can see why it is true when it is positive, but it looks false otherwise.Log in to reply

– Daniel Liu · 2 years, 9 months ago

Right, the inequality only holds for positive reals. Thanks for reminding me, I'll fix it.Log in to reply

– Beakal Tiliksew · 2 years, 9 months ago

Prove \[\lfloor 3x\rfloor = \lfloor x\rfloor + \left\lfloor x+\frac 13 \right\rfloor + \left\lfloor x+\frac 23 \right\rfloor \]Log in to reply

– Daniel Liu · 2 years, 9 months ago

That is just Hermite's Identity for \(n=3\).Log in to reply

– Daniel Remo · 2 years, 9 months ago

I have seen an equivalent identity to Hermite's but for ceiling functions: do you know by any chance if there is a separate name for it to quote in olympiads, or whether I'd need to prove it?Log in to reply

It is not hard to prove from Hermite's Identity. there is no name for it, so you are better off proving it. – Daniel Liu · 2 years, 9 months ago

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– Finn Hulse · 2 years, 9 months ago

Hey yeah I realized that too. :DLog in to reply

– Sharky Kesa · 2 years, 9 months ago

That's what I did. I was gong to post a hint.Log in to reply

– Sharky Kesa · 2 years, 9 months ago

I'll post it anyway.Log in to reply

– Finn Hulse · 2 years, 9 months ago

Mmm. Essentially, you're boiling down my solution and going backwards. Cool! :DLog in to reply

@Finn Hulse @Daniel Liu – Mardokay Mosazghi · 2 years, 9 months ago

I am so bad at proofs? Help me how can i practice and where should i start.Log in to reply

– Sharky Kesa · 2 years, 9 months ago

Brilliant. Solve some problems and read their solutions. You learn loads of stuff from well-written solutions. When I started Brilliant, I didn't even know what floor and ceiling functions. Now I can solve these sorts of proof equations mentally.Log in to reply

– Calvin Lin Staff · 2 years, 9 months ago

Good for you!Log in to reply

– Finn Hulse · 2 years, 9 months ago

Excuse me can we talk?Log in to reply

– Finn Hulse · 2 years, 9 months ago

I'm not too great either. I tend to draw mine out and make them longer than they need to be, because I have trouble explaining really simple concepts... It's weird. :PLog in to reply

– Sharky Kesa · 2 years, 9 months ago

Experience helps.Log in to reply

I Think the Answer is 56 – Henry Chen · 2 years, 9 months ago

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