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Prove that for $a,b,c > 0$

$$\frac{a+b-2c}{b+c}+\frac {b+c-2a}{c+a}+\frac {c+a-2b}{a+b} >0$$

What I did is this:- Let $f(a,b,c)=\frac{a+b-2c}{b+c}+\frac {b+c-2a}{c+a}+\frac {c+a-2b}{a+b}$. Therefore

$$\begin{align}f(a,b,c)&=\frac{a+b-2c}{b+c}+\frac {b+c-2a}{c+a}+\frac {c+a-2b}{a+b}\\&=(a+b+c)\left(\frac{1}{b+c}+\frac {1}{c+a}+\frac {1}{a+b}\right)-3\left(\frac{c}{b+c}+\frac {a}{c+a}+\frac {b}{a+b}\right)\end{align}$$

The first term is greater than $9/2$ by the Cauchy-Schwarz inequality. How do I prove that the term inside the second bracket is less than $3/2$? Can I use the rearrangement inequality here??

Daniel Fischer
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Abhishek Bakshi
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    $a,b,c$ must all be different, yes? – user31415926535 Dec 29 '14 at 13:35
  • You can't prove it like that. When $a\to 0$, $b=1$, and $c\to\infty$, we have $\frac c{b+c}+\frac a{c+a}+\frac b{a+b}\to 1+0+1=2$. – user2345215 Dec 29 '14 at 13:41
  • Why is this tagged ([tag:linear-algebra])? – Martin Sleziak Dec 29 '14 at 15:09
  • @MartinSleziak i may be wrong but i did it from a book of algebra, hence i thought that linear algebra would be a suitable tag.. – Abhishek Bakshi Dec 29 '14 at 17:35
  • If the source is a book about algebra (and not specifically linear algebra) it would suggest ([tag:algebra-precalculus]) or ([tag:abstract-algebra]) as a tag. But my opinion is that leaving the question only with ([tag:inequality]) tag seems to be quite reasonable. – Martin Sleziak Dec 29 '14 at 17:38
  • I worked out an alternate way of proving the inequality without using the identity in answer, let me know if you are interested :-) – r9m Dec 29 '14 at 21:18
  • @r9m yes, of course...why don't you post it as an answer...and also i am selected to give the Indian national mathematics olympiad this coming february....so could you give me some good algebra and number theory problems to solve??? – Abhishek Bakshi Dec 30 '14 at 10:18
  • @AbhishekBakshi see comment and the linked solution :) – r9m Jan 09 '15 at 07:36

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your inequality is equivalent to $${\frac {{a}^{3}+{a}^{2}b-2\,{a}^{2}c-2\,a{b}^{2}+a{c}^{2}+{b}^{3}+{b}^ {2}c-2\,b{c}^{2}+{c}^{3}}{ \left( b+c \right) \left( c+a \right) \left( a+b \right) }} >0$$ and the numerator can written as $$a(a-c)^2+b(b-a)^2+c(b-c)^2\geq 0$$ and the equal sign holds if $$a=b=c$$

Dr. Sonnhard Graubner
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