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August 9th, 2014, 04:52 PM   #1
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good proof

a>0;b>0 ;c>0
1/a+1/b+1/c>=a+b+c
proof that
a+b+c>=abc
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August 10th, 2014, 12:37 AM   #2
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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 10th, 2014 at 12:38 AM. Reason: Oops
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August 10th, 2014, 02:20 PM   #3
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Quote:
Originally Posted by i8sumpi View Post
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.
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August 10th, 2014, 07:19 PM   #4
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$\displaystyle a>0; \, b>0;\, c>0$

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 $\displaystyle abc\le a+b+c$
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August 11th, 2014, 07:11 AM   #5
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Quote:
Originally Posted by neelmodi View Post
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?
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August 11th, 2014, 07:32 AM   #6
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Quote:
Originally Posted by Evgeny.Makarov View Post
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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