<s>[INST] If $a,b,c>0\;,$ Then value of $\displaystyle \lfloor \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\rfloor $, 
If $a,b,c>0\;,$ Then value of $\displaystyle \bigg \lfloor \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\bigg\rfloor $, Where $\lfloor x \rfloor $ Rep. floor function of $x$.

$\bf{My\; Try::}$ Using $b+c>a$ and $c+a>b$ and $a+b>c$
So $\displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}<3$
And Using $\bf{A.M\geq H.M}\;,$ we get
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}>\frac{9}{\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}}$$
I did not Understand How can I calculate its Lower bond, Help Required
Thanks
 [/INST] Hint a small trick is that if you take $3$ numbers greater than $0$ you will always have ywo such pairs whose sum is greater than $3$rd number so division of two pairs will give answer of division as $0<x<1$ thus the sum would always be less than $2$ implying iys sums floor function would be $1$.
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