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 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
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Quote:
 Originally Posted by i8sumpi 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 >=
You didn't confirm it. If it is false, then you can show it is false with just
one counterexample.

 August 10th, 2014, 06:19 PM #4 Senior Member   Joined: Aug 2014 From: United States Posts: 137 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
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Quote:
 Originally Posted by neelmodi Thus $\displaystyle abc\le a+b+c$
How do you derive this from $\displaystyle ab+bc+ac\ge abc(a+b+c)$ and $\displaystyle \sqrt[3]{abc}\ge \displaystyle \frac{a+b+c} 3$? Also, shouldn't the AM-GM inequality be reversed?

August 11th, 2014, 06:32 AM   #6
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Quote:
 Originally Posted by Evgeny.Makarov How do you derive this from $\displaystyle ab+bc+ac\ge abc(a+b+c)$ and $\displaystyle \sqrt[3]{abc}\ge \displaystyle \frac{a+b+c} 3$? Also, shouldn't the AM-GM inequality be reversed?
Oh sorry, my bad... then that changes things.

Okay, please disregard my attempt at a proof.

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