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Prove that $\lfloor 2x \rfloor + \lfloor 2y \rfloor \geq \lfloor x \rfloor + \lfloor y \rfloor + \lfloor x+y \rfloor$ for all real $x$ and $y$.

How should I solve this? I can't think of a way with casework and I can't really simplify it more. Thanks in advance for posting a proof!

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closed as off-topic by heropup, Daniel W. Farlow, Parcly Taxel, user91500, R_D Sep 12 '16 at 4:55

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – heropup, Daniel W. Farlow, Parcly Taxel, user91500, R_D
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level. $\endgroup$ – The Great Duck Sep 11 '16 at 23:47
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With these propositions, the trick is usually to write $x=k+e$ and $y=n+f$ where $n$ and $k$ are integers and $0\leq e,g <1$. Then your left side is $2k+\lfloor 2e\rfloor + 2n + \lfloor 2f \rfloor$. And your right side is $k+n+(k+n)+\lfloor e+f \rfloor.$ Cancelling, you need to show now that $\lfloor 2e\rfloor +\lfloor 2f \rfloor \geq \lfloor e+f \rfloor.$ You can do this by considering 4 cases, depending on whether $e$ and $f$ are less than or as big as $1/2$. E.g. if $0\leq e <1/2$ and $1/2\leq f <1$ then the left side is at least 1 and the right side is at most 1.

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  • $\begingroup$ Please do not answer homework questions. $\endgroup$ – The Great Duck Sep 18 '16 at 23:38
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It suffices to consider $0\le x\le y<1$. The right hand side is $\le 1$, and it is $>0$ only if $x+y\ge 1$, which requires $y\ge \frac 12$, which makes the left $\ge1$.

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  • $\begingroup$ Please do not answer homework questions. $\endgroup$ – The Great Duck Sep 18 '16 at 23:39
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In another way, premised that: $$ \begin{gathered} \left\lfloor {x + y} \right\rfloor = \left\lfloor {\left\lfloor x \right\rfloor + \left\{ x \right\} + \left\lfloor y \right\rfloor + \left\{ y \right\}} \right\rfloor = \hfill \\ = \left\lfloor x \right\rfloor + \left\lfloor y \right\rfloor + \left\lfloor {\left\{ x \right\} + \left\{ y \right\}} \right\rfloor = \hfill \\ = \left\lfloor x \right\rfloor + \left\lfloor y \right\rfloor + \left[ {1 \leqslant \left\{ x \right\} + \left\{ y \right\}} \right] \hfill \\ \end{gathered} $$ where the square brackets indicate the Iverson bracket ($\left[ {FALSE} \right] = 0,\;\left[ {TRUE} \right] = 1$)
then $$ \begin{gathered} \left\lfloor {2x} \right\rfloor + \left\lfloor {2y} \right\rfloor = 2\left\lfloor x \right\rfloor + 2\left\lfloor y \right\rfloor + \left[ {1 \leqslant 2\left\{ x \right\} + 2\left\{ y \right\}} \right] \hfill \\ \left\lfloor x \right\rfloor + \left\lfloor y \right\rfloor + \left\lfloor {x + y} \right\rfloor = 2\left\lfloor x \right\rfloor + 2\left\lfloor y \right\rfloor + \left[ {1 \leqslant \left\{ x \right\} + \left\{ y \right\}} \right] \hfill \\ \end{gathered} $$ and, clearly $$ \left[ {1 \leqslant \left\{ x \right\} + \left\{ y \right\}} \right] \leqslant \left[ {1/2 \leqslant \left\{ x \right\} + \left\{ y \right\}} \right] $$

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  • $\begingroup$ Please do not answer homework questions. $\endgroup$ – The Great Duck Sep 18 '16 at 23:39
  • $\begingroup$ @TheGreatDuck what do you actually mean to say? most of the post originate from homework questions, and as far as I understood about the policy of this site, help is due when it is required in a "non-totally passive " way, which did not appeared to me to be case here. $\endgroup$ – G Cab Sep 19 '16 at 9:13
  • $\begingroup$ It was closed as off topic therefore your answer violates the rules. Off topic questions shouldn't be answered. $\endgroup$ – The Great Duck Sep 21 '16 at 21:19

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