Inequality $\frac{b}{a}+\frac{c}{b}+\frac{a}{c}-\frac{c}{a+b}-\frac{a}{b+c}-\frac{b}{a+c}\ge 3/2$ 
prove that for $a,b,c$ being positives and $a+b+c=1$:$$\frac{b}{a}+\frac{c}{b}+\frac{a}{c}-\frac{c}{a+b}-\frac{a}{b+c}-\frac{b}{a+c}\ge 3/2$$

This is a very interesting inequality which i came upon accidentally.We also see that the condition $a+b+c=1$ is not needed.I modified the inequality slightly using Nesbitt's inequality and hence it boils down to prove an even stronger inequality $$\sum_{cyc}\left(\frac{b}{a}-\frac{2c}{a+b}\right)\ge 0$$ Since this was a stronger version i checked WA which shows that it is valid.I have tried to get an SOS but failed.The problem is that even after complete expanding and cross-multiplying we get a cyclic inequality and hence Muirheads theorem fails.

First inequality:
$$ LHS = \sum_{cyc} \dfrac{bc}{a(a+c)}  = \sum_{cyc} \dfrac{(bc)^2}{a^2bc+abc^2} \ge \dfrac{(ab+bc+ca)^2}{ 2abc(a+b+c)} \ge \dfrac{3}{2}$$
