The question: If $a,b,c$ are the side lengths of a triangle, then
$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}<2$
I added $3$ to both sides:
$\frac{a}{b+c}+1+\frac{b}{c+a}+1+\frac{c}{a+b}+1<5$
Equivalent to $(a+b+c)(\sum \frac{(c+a)(a+b)}{(c+a)(a+b)(b+c)})<5$
Equivalent to $\frac{(a+b+c)^3}{(c+a)(a+b)(b+c)}<5$
$LHS<\frac{\sum a^3}{\sum a^2(b+c)}+\frac{3\sum a^2(b+c)}{\sum a^2(b+c)}+\frac{6abc}{(b+c)(c+a)(a+b)}$
$<1+3+\frac{6abc}{8\sqrt(bccaab)}=19/4$
by AM-GM
For some reason I find it really odd that it doesn't actually have the same bound as the Q. Is there any mistake?
One error is here: $$ \sum(c+a)(a+b)=a^2+b^2+c^2+3ab+3bc+3ca\ne(a+b+c)^2. $$
In any case the solution is unnecessary complicated. Assume without loss of generality $a\le b\le c$. Then: $$ \frac a{b+c}+\frac b{c+a}+\frac c{a+b}\le\frac a{a+b}+\frac b{a+b}+\frac c{a+b} =1+\frac c{a+b}<2. $$