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# I need help fast. Questions due very soon. Please hurry. Thanks!

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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.

Please provide a full solution. Thanks!

Oct 18, 2018

#6
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My attempt at a solution:

.

Oct 18, 2018

#1
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$$\left |x \right | + \left |y \right | = \left |x \right | +\left |y \right |$$(1)

We know:

$$\left |x \right | + \left | y \right |$$>$$\left | x+y \right |$$ (2)

We (1)+(2)

$$\left |2x \right | + \left | 2y\right |$$ >$$\left | x \right | + \left | y \right | +\left | x+y \right |$$

We prove it!

Hope it helps!

Oct 18, 2018
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This is not what the guest wants to prove, I think you confused the floor function with the absolute value function

Guest Oct 18, 2018
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Oh now i see it.It's 8:57 a.m. and im awake all night I was studying.I'm sorry!

Thanks for the observation!

Dimitristhym  Oct 18, 2018
edited by Dimitristhym  Oct 18, 2018
#4
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We all make mistakes

Guest Oct 18, 2018
#5
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I'd like to see an answer to this question too.

Oct 18, 2018
#6
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My attempt at a solution:

.

Alan Oct 18, 2018
#7
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Thanks Alan, I like that logic :)

I just got in a mess when I tried to solve it.

Melody  Oct 18, 2018
#8
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THANK YOU SOOOO MUCH GUYS! I really appreciate this. Thanks!

Side note: hmmm, i should probably make an account.

Side Side note: Oh, and my name is also Alan spelled the same way. Hehe

Oct 18, 2018
edited by Guest  Oct 18, 2018
#9
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Yes, you should make an account and become known to us.  :)

Melody  Oct 19, 2018
#10
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I did! My username is now FencingKat

FencingKat  Oct 22, 2018