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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

Best Answer 

 #6
avatar+27547 
+5

My attempt at a solution:

 

.

 Oct 18, 2018
 #1
avatar+316 
+1

\(\left |x \right | + \left |y \right | = \left |x \right | +\left |y \right |\)(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
 #2
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0

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
 #3
avatar+316 
0

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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+1

We all make mistakes smiley

Guest Oct 18, 2018
 #5
avatar+99327 
0

I'd like to see an answer to this question too.   frown

 Oct 18, 2018
 #6
avatar+27547 
+5
Best Answer

My attempt at a solution:

 

.

Alan Oct 18, 2018
 #7
avatar+99327 
+1

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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+1

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
avatar+99327 
+2

Yes, you should make an account and become known to us.  :)

Melody  Oct 19, 2018
 #10
avatar+39 
0

I did! My username is now FencingKat

FencingKat  Oct 22, 2018

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