<s>[INST] How to prove $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} <2$? Prove the inequality for a triangle with sides $a,b,c$ we have $$\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b} <2$$
Trial: Since $a,b,c$ are sides of a triangle I know  $a+b>c,b+c>a,a+c>b$  
 [/INST] \begin{align*}\frac a{b+c}+\frac b{c+a}+\frac c{a+b}<2&\iff 3-2<1-\frac a{b+c}+1-\frac b{c+a}+1-\frac c{a+b}\\&\iff 1<\frac {b+c-a}{b+c}+\frac {c+a-b}{c+a}+\frac {a+b-c}{a+b}\end{align*}
Now
$$\frac {b+c-a}{b+c}+\frac {c+a-b}{c+a}+\frac {a+b-c}{a+b}>\frac {b+c-a}{a+b+c}+\frac {c+a-b}{a+b+c}+\frac {a+b-c}{a+b+c}=1$$
You need the triangle inequality to ensure the fractions on the left side are positive, so you decrease them by increasing the denominator.
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