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I was trying to prove $\lceil \sqrt{2}n \rceil + \lceil \sqrt{2}m \rceil \geq \lceil \sqrt{2}(n+m) \rceil$ where $m,n\in \mathbb{z}$

Direct proof I tried but could not figure out. I tried fixing m and induction on n, for n = 1 holds,

then i got stucked for k+1 case.

Is there anything fancy required here which I may have to consider?

Thanks.

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Simply show that for all real numbers $a$ and $b$, we have

$$\lceil a \rceil + \lceil b \rceil \geq \lceil a+b \rceil.$$

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  • $\begingroup$ This is a nice example of a case where it is easier to prove a more general theorem than a more specific one. $\endgroup$ – MJD Mar 30 '14 at 20:36
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This can be proven directly from properties of the ceiling function without much bother. Think about how the ceiling function distributes over addition.

Hint for a clever (but unnecessarily difficult) solution: consider a sequence of cleverly chosen rational numbers greater than $\sqrt{2}$ that converge and then prove the inequality. For those numbers.

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