for triamgle $ABC$ prove that $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} <2$ [duplicate]

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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

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

• Now substitute the values of x, y, z and you will get the answer. – Wishwas Oct 27 '16 at 8:06
• 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!! – tatan Oct 28 '16 at 7:19

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 $\frac{a}{a+b}+\frac{b}{c+a}+\frac{c}{a+b}<2$
• We have $\frac{a}{a+b}\ge \frac{a}{a+c}$, because $b\le c$. You made a mistake. – user26486 Oct 5 '17 at 19:18