Simple Nearest Integer Inequality

Let x,yx,y be positive real numbers. Prove that x+yx+yx+yx+y\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 x1,x2xnx_1,x_2\ldots x_n, that i=1nxii=1nxii=1nxii=1nxi\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}

Note by Daniel Liu
5 years ago

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Cool problem! I think this is a valid approach:

Looking at the first equation, we see that equality is attained if xx and yy are both integers, which would make the equation just 111 \leq 1. Split this into two cases:

Case 1:

x+y<x+y\lfloor{x}\rfloor+\lfloor{y}\rfloor < \lfloor{x+y}\rfloor. In this case, obviously neither xx or yy are integers otherwise their sum would be the same as the floor of their sum. Thus the numerator of the RHS will become x+1+y+1\lfloor{x}\rfloor+1+\lfloor{y}\rfloor+1 and denominator x+y+1\lfloor{x}\rfloor+\lfloor{y}\rfloor+1. Obviously the LHS will be <1<1 and the RHS >1>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<1 and the denominator >1>1 and the reverse on the RHS.

Case 2:

Both xx and yy are integers. The inequality becomes

x+yx+yx+yx+y\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:

x+y=x+y\lfloor{x}\rfloor+\lfloor{y}\rfloor = \lfloor{x+y}\rfloor. This will be true as long as the fractional part of xx and yy do not have a sum >1>1, unlike Case 1. So, like, xx could be 3.33.3 and yy could be 1.61.6, and 0.6+0.310.6+0.3 \ngeq 1. In this case, the inequality becomes

x+yfrac(x)frac(y)x+yfrac(x)frac(y)x+y+1frac(x)frac(y)x+y+1frac(x)frac(y)\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 frac(n)\text{frac}(n) represents the fractional part, or just nnn-\lfloor{n}\rfloor. Obviously, both sides are equal and the inequality is satisfied. We see that for all x1,x2,x3,x_1, x_2, x_3, \dots this is satisfied as well since there will always be that extra +1+1 in the numerator and denominator of the RHS since the sum of the fractional parts is <1<1.

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

Quod erat demonstratum.

Finn Hulse - 5 years ago

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Simpler:

Let x=x1+xfx=x_1+x_f and y=y1+yfy=y_1+y_f where x1,y1x_1,y_1 are integers and xf,yfx_f,y_f are the fractional part of x,yx,y respectively.

Note that x+yx+y\lfloor x\rfloor +\lfloor y\rfloor \le \lfloor x+y\rfloor because x1+y1x1+y1+xf+yfx_1+y_1 \le x_1+y_1+\lfloor x_f+y_f\rfloor

This equation becomes x+yx+y1\dfrac{\lfloor x\rfloor +\lfloor y\rfloor}{\lfloor x+y\rfloor}\le 1

Also, note that x+yx+y\lceil x \rceil +\lceil y \rceil \ge \lceil x+y\rceil because x1+1+y1+1x1+y1+xf+yfx_1+1+y_1+1\ge x_1+y_1+\lceil x_f+y_f\rceil

This equation becomes x+yx+y1\dfrac{\lceil x \rceil +\lceil y \rceil}{\lceil x+y\rceil}\ge 1

Putting the two equations together, we have x+yx+y1x+yx+y\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 - 5 years ago

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Mmm. Essentially, you're boiling down my solution and going backwards. Cool! :D

Finn Hulse - 5 years ago

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@Finn Hulse I am so bad at proofs? Help me how can i practice and where should i start.@Finn Hulse @Daniel Liu

Mardokay Mosazghi - 5 years ago

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@Mardokay Mosazghi 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.

Sharky Kesa - 5 years ago

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@Sharky Kesa Good for you!

Calvin Lin Staff - 5 years ago

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@Calvin Lin Excuse me can we talk?

Finn Hulse - 5 years ago

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@Mardokay Mosazghi 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. :P

Finn Hulse - 5 years ago

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@Finn Hulse Experience helps.

Sharky Kesa - 5 years ago

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

Sharky Kesa - 5 years ago

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@Sharky Kesa I'll post it anyway.

Sharky Kesa - 5 years ago

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Prove 3x=x+x+13+x+23\lfloor 3x\rfloor = \lfloor x\rfloor + \left\lfloor x+\frac 13 \right\rfloor + \left\lfloor x+\frac 23 \right\rfloor

Beakal Tiliksew - 5 years ago

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@Beakal Tiliksew That is just Hermite's Identity for n=3n=3.

Daniel Liu - 5 years ago

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@Daniel Liu Hey yeah I realized that too. :D

Finn Hulse - 5 years ago

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@Daniel Liu 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?

Daniel Remo - 5 years ago

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@Daniel Remo Indeed, i=0n1x+in=nx+n1\sum_{i=0}^{n-1}\left\lceil x+\dfrac{i}{n}\right\rceil = \lceil nx\rceil +n-1

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 - 5 years ago

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How can you claim that x+yx+y    x+yx+y1\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 x+y\lfloor x+y\rfloor can be negative? I can see why it is true when it is positive, but it looks false otherwise.

mathh mathh - 5 years ago

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@Mathh Mathh Right, the inequality only holds for positive reals. Thanks for reminding me, I'll fix it.

Daniel Liu - 4 years, 11 months ago

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Hint:

nnn+1\lfloor{n}\rfloor \leq \lceil{n}\rceil \leq \lfloor{n}\rfloor + 1

Sharky Kesa - 5 years ago

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I think the << should be changed to a \leq.

Finn Hulse - 5 years ago

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No, cos it isn't possible.

Sharky Kesa - 5 years ago

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@Sharky Kesa 3.5=3.5+14=3+1\lceil{3.5}\rceil=\lfloor{3.5}\rfloor+1 \longrightarrow 4=3+1.

Finn Hulse - 5 years ago

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@Finn Hulse Sorry, didn't realise.

Sharky Kesa - 5 years ago

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@Sharky Kesa It's okay bro. I like your spelling of "realize". :D

Finn Hulse - 5 years ago

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@Finn Hulse 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.

Sharky Kesa - 5 years ago

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@Sharky Kesa Yeah bro, it's okay I know. Wait do you write cursive? :O

Finn Hulse - 5 years ago

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@Finn Hulse Yes. We have to write in cursive/running writing. It's faster.

Sharky Kesa - 5 years ago

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I Think the Answer is 56

Henry Chen - 4 years, 12 months ago

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