<s>[INST] Finding the range of $\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b}$, for $a$, $b$, $c$ the sides of a triangle 
If $a$, $b$, and $c$ are the three sides of a triangle, then 
  $$\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b}$$ 
  lies in what interval?

 [/INST] For a $\triangle ABC,$ we have $a+b>c$ and $b+c>a$ and $c+a>b$
For $\bf{Upper\; bound}$
$$a+b>c\Rightarrow 2(a+b)>a+b+c\Rightarrow \frac{1}{2(a+b)}<\frac{1}{a+b+c}\Rightarrow \frac{c}{a+b}<\frac{2c}{a+b+c}$$
In a Similar way for other two expression, And then adding , We get
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}<\frac{2a}{a+b+c}+\frac{2b}{a+b+c}+\frac{2c}{a+b+c} = 2$$
Similarly For $\bf{Lower\; bound}$
$$a>0\Rightarrow a+b+c>b+c\Rightarrow \frac{1}{a+b+c}<\frac{1}{b+c}\Rightarrow \frac{a}{a+b+c}<\frac{a}{b+c}$$
Similar way for other two expression, And then Adding, We get
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c} = 1$$
So we get $$1<\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}<2$$
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