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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TopNewestHint:

\[\lfloor{n}\rfloor \leq \lceil{n}\rceil \leq \lfloor{n}\rfloor + 1\]

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

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No, cos it isn't possible.

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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 \(x\) 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.

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Simpler:

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. \(\Box\)

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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.

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

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It is not hard to prove from Hermite's Identity. there is no name for it, so you are better off proving it.

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

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

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@Finn Hulse @Daniel Liu

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

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I Think the Answer is 56

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