August 9th, 2014, 03:52 PM  #1 
Senior Member Joined: Nov 2013 Posts: 434 Thanks: 8  good proof
a>0;b>0 ;c>0 1/a+1/b+1/c>=a+b+c proof that a+b+c>=abc 
August 9th, 2014, 11:37 PM  #2 
Newbie Joined: Aug 2014 From: Earth Posts: 4 Thanks: 0 Math Focus: Cartesian plane and set theory 
Let me try a = 1 b = 2 c = 3 1/1 + 1/2 + 1/3 3/6 + 2/6 = 5/6 1+2+3 = 6 1x2x3 = 6 Yep, can confirm good proof there Edit: minus the 1/a part, it should be =< not >= Last edited by i8sumpi; August 9th, 2014 at 11:38 PM. Reason: Oops 
August 10th, 2014, 01:20 PM  #3 
Banned Camp Joined: Jun 2014 From: Earth Posts: 945 Thanks: 191  
August 10th, 2014, 06:19 PM  #4 
Senior Member Joined: Aug 2014 From: United States Posts: 136 Thanks: 21 Math Focus: Learning 
$\displaystyle a>0; \, b>0;\, c>0$ If $\displaystyle \displaystyle \frac{1} {a} +\frac{1} {b} +\frac{1} {c}\ge a+b+c$ then $\displaystyle ab+bc+ac\ge abc(a+b+c)$ (I just multiplied both sides by abc) You can use the fact that the geometric mean is greater than or equal to the arithmetic mean. $\displaystyle \sqrt[3]{abc}\ge \displaystyle \frac{a+b+c} 3 $ Thus $\displaystyle abc\le a+b+c$ 
August 11th, 2014, 06:11 AM  #5 
Senior Member Joined: Dec 2013 From: Russia Posts: 327 Thanks: 108  
August 11th, 2014, 06:32 AM  #6  
Senior Member Joined: Aug 2014 From: United States Posts: 136 Thanks: 21 Math Focus: Learning  Quote:
Okay, please disregard my attempt at a proof.  

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