Inequality based on triangle: $\frac{3}{2}\le\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}<2$ 
If $a,b,c$ are sides of a triangle, prove that ${3\over2 }\le {{a\over b+c}} + {{b\over c+a}}+{{c\over a+b}} \lt 2$ .


Variational Approach
Without loss of generality, assume that $a+b+c=1$. Then
$$
\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}\tag{1}
$$
has an interior extreme when
$$
\frac{\delta a}{(1-a)^2}+\frac{\delta b}{(1-b)^2}+\frac{\delta c}{(1-c)^2}=0\tag{2}
$$
for any variation so that
$$
\delta a +\delta b+\delta c=0\tag{3}
$$
Linearity says that we need $a=b=c=\frac13$, so that
$$
\bbox[5px,border:2px solid #C0A000]{\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}=\frac32}\tag{4}
$$

Due to symmetry and the triangle inequality, an edge extreme occurs only when $c=a+b=\frac12$ or when $c=0$, in which case $a+b=1$. In either case, we are finding an extreme of
$$
\frac{a}{1-a}+\frac{b}{1-b}\tag{5}
$$
An interior extreme of $(5)$ occurs when
$$
\frac{\delta a}{(1-a)^2}+\frac{\delta b}{(1-b)^2}=0\tag{6}
$$
for any variation so that
$$
\delta a +\delta b=0\tag{7}
$$
Linearity says that we need $a=b$, so that if $c=\frac12$, then $a=b=\frac14$ and
$$
\bbox[5px,border:2px solid #C0A000]{\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}=\frac53}\tag{8}
$$
or if $c=0$, then $a=b=\frac12$ and
$$
\bbox[5px,border:2px solid #C0A000]{\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}=2}\tag{9}
$$

We don't need to worry about the edge edge case where $a=0$ or $b=0$ since those cases are covered by $(9)$. Therefore, the possible extremes are  given in $(4)$, $(8)$, and $(9)$. Thus,
$$
\bbox[5px,border:2px solid #C0A000]{\frac32\le\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}\le2}\tag{10}
$$
