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This question already has an answer here:

Let $ a,b,c$ be lengths of sides triangle $ ABC$. Prove that: $\displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} <2$

$\bf{My\; Try::}$ For a $\triangle $ with sides $a,b,c$

$a+b>c$ and $b+c>a$ and $a+b>c$

Now let $x=a+b-c>0$ and $y=b+c-a>0$ and $z=c+a-b>0$

So $\displaystyle 2a=x+z$ and $2b=x+y$ and $2c=y+z$

So equality convert into $$\frac{x+z}{x+2y+z}+\frac{x+y}{x+y+2z}+\frac{y+z}{2x+y+z}<2$$

How can i solve above inequality, Help required, Thanks

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marked as duplicate by Joey Zou, Parcly Taxel, lab bhattacharjee, Claude Leibovici, Community Oct 27 '16 at 8:44

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Now substitute the values of x, y, z and you will get the answer. $\endgroup$ – Wishwas Oct 27 '16 at 8:06
  • $\begingroup$ This was a famous International Olympiad problem....the whole question was to prove that $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}$ is bounded between $2$ and $\frac 32$...For the $\frac 32$ part check out Nesbitt's Inequality-en.wikipedia.org/wiki/Nesbitt%27s_inequality .....Thanks and hope this helps!! $\endgroup$ – tatan Oct 28 '16 at 7:19
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By symmetry we may assume that $a\leq b\leq c$ . In that case

$\frac {a}{a+b}+\frac {b}{c+a}+\frac{c}{a+b}\leq \frac{a}{a+c}+\frac{c}{c+a}+\frac {c}{a+b}=1+\frac{c}{a+b}$

but $1+\frac {c}{a+b}<2$

so $\frac{a}{a+b}+\frac{b}{c+a}+\frac{c}{a+b}<2$

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  • $\begingroup$ The question is a duplicate so we already have answers for this. Look at math.stackexchange.com/questions/1903775/… $\endgroup$ – tatan Oct 27 '16 at 8:56
  • $\begingroup$ We have $\frac{a}{a+b}\ge \frac{a}{a+c}$, because $b\le c$. You made a mistake. $\endgroup$ – user26486 Oct 5 '17 at 19:18

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