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S Mar 11, 2017 at 20:15 history suggested Xam CC BY-SA 3.0
Improved formatting
Mar 11, 2017 at 19:48 review Suggested edits
S Mar 11, 2017 at 20:15
Mar 11, 2017 at 5:38 vote accept MAEA2
Mar 11, 2017 at 5:32 answer added Jeremy Rouse timeline score: 40
Mar 11, 2017 at 4:49 history edited MAEA2 CC BY-SA 3.0
added 18 characters in body
Mar 11, 2017 at 4:47 comment added MAEA2 @T.Amdeberhan Oh thanks! Now, I want to find the other answer which fulfils $a,b,c>1$
Mar 11, 2017 at 4:39 comment added T. Amdeberhan On the trivial side, $(3,1,1)$ is a solution.
Mar 11, 2017 at 3:41 comment added MAEA2 @KConrad I think $\mathbb{N}=1,2,3,\ldots$.
Mar 11, 2017 at 3:31 history edited Gottfried Helms CC BY-SA 3.0
spelling in header
Mar 11, 2017 at 3:27 comment added KConrad I guess you don't regard $0$ as a natural number? Otherwise let one of $a, b, c$ be $0$ and the other two be $1$.
Mar 11, 2017 at 3:27 comment added JMP you can edit your own questions if you make a mistake - look just right and down of the vote arrows
Mar 11, 2017 at 3:25 comment added MAEA2 sorry,it's first time I use math over flow. Thanks for approvemenrt!
S Mar 11, 2017 at 3:23 history suggested JMP CC BY-SA 3.0
changed formula as per comment by author
Mar 11, 2017 at 3:22 review Suggested edits
S Mar 11, 2017 at 3:23
Mar 11, 2017 at 3:12 comment added MAEA2 sorry, $\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}$ is correct
Mar 11, 2017 at 3:11 review First posts
Mar 11, 2017 at 3:22
Mar 11, 2017 at 3:09 history asked MAEA2 CC BY-SA 3.0