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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## Comments

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

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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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$n=3$.

That is just Hermite's Identity forLog in to reply

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$\sum_{i=0}^{n-1}\left\lceil x+\dfrac{i}{n}\right\rceil = \lceil nx\rceil +n-1$

Indeed,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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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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Hint:

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

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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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$\lceil{3.5}\rceil=\lfloor{3.5}\rfloor+1 \longrightarrow 4=3+1$.

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I Think the Answer is 56

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