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 ${3\over2 }\le {{a\over b+c}} + {{b\over c+a}}+{{c\over a+b}} \lt 2$ .

 A: 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,
$$
\bbox[5px,border:2px solid #C0A000]{\frac32\le\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}\le2}\tag{10}
$$
A: The left inequality.
$\sum\limits_{cyc}\frac{a}{b+c}-\frac{3}{2}=\sum\limits_{cyc}\left(\frac{a}{b+c}-\frac{1}{2}\right)=\sum\limits_{cyc}\frac{a-b-(c-a)}{2(b+c)}=\sum\limits_{cyc}\frac{(a-b)^2}{2(a+c)(b+c)}\geq0$.
The right inequality.
$\sum\limits_{cyc}\frac{a}{b+c}<\sum\limits_{cyc}\frac{a+a}{a+b+c}=2$
A: The left inecuality:
Use Titu's Lemma and $a^2+b^2+c^2\ge ab+bc+ca$
$$\frac a{b+c}+\frac b{c+a}+\frac c{a+b}=\frac {a^2}{ab+ac}+\frac {b^2}{bc+ba}+\frac {c^2}{ca+bc}\ge$$
$$\ge\frac{(a+b+c)^2}{2(ab+bc+ca)}=\frac{a^2+b^2+c^2}{2(ab+bc+ca)}+1\ge\frac12+1=\frac32$$
The left inecuality is Nesbitt's inequality
A: We show the right side of the inequality, that is,
\begin{align*}
\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} < 2, 
\end{align*}
which is equivalent to 
\begin{align*}
\frac{b+c-a}{b+c}+\frac{c+a-b}{c+a}+\frac{a+b-c}{a+b} > 1.
\end{align*}
Let $x=b+c-a$, $y=c+a-b$, and $z=a+b-c$. Then $x>0$, $y>0$, and $z>0$. Moreover,
\begin{align*}
\frac{b+c-a}{b+c}+\frac{c+a-b}{c+a}+\frac{a+b-c}{a+b} &= \frac{x}{x+\frac{y+z}{2}}+\frac{y}{y+\frac{x+z}{2}}+\frac{z}{z+\frac{x+y}{2}}\\
&=\frac{x^2}{x^2+x\frac{y+z}{2}}+\frac{y^2}{y^2+y\frac{x+z}{2}}+\frac{z^2}{z^2+z\frac{x+y}{2}}\\
&\ge\frac{(x+y+z)^2}{x^2+y^2+z^2+xy+xz+yz} > 1.
\end{align*}
We are done.
