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.
 A: First inequality:
$$ LHS = \sum_{cyc} \dfrac{bc}{a(a+c)}  = \sum_{cyc} \dfrac{(bc)^2}{a^2bc+abc^2} \ge \dfrac{(ab+bc+ca)^2}{ 2abc(a+b+c)} \ge \dfrac{3}{2}$$
A: Using a computer algebra system, the inequality in the title turns out to be equivalent after expanding to the inequality:
$$
2\sum_{\text{cyclic}}a^4b^2
+
2\sum_{\text{cyclic}}a^3b^3
\ge
\sum_{\text{cyclic}}a^3bc(b+c)
+ 6a^2b^2c^2\ .
$$
Now we represent the monomial powers $(r,s,t)$ in the plane $r+s+t=6$, and in each "node" that appears in the inequality we place the corresponding coefficient. This makes it easier to find a domination scheme. The picture is as follows:
                b^6
                .
              .  .
            .   .  2
          2  -1  -1  2
        2  -1  -6  -1  .
      .   .  -1  -1  .  .
    .   .   .   2  2  .  . 
a^6                        c^6

A comment on the scheme. The "extremal vertices" are marked with $a^6$ alias $(6,0,0)$, $b^6$ alias $(0,6,0)$, and $c^6$ alias $(0,0,6)$.
Consider the "base line" joining the vertices for $b^6$ and $c^6$. Parallels to this line passing through the lattice points are lines with constant $a$-part in the monomial. So the parallel line "immediately after the line from $b^6$ to $c^6$" is the line from $ab^5$ to $ac^5$, and the lattice points on it are corresponding to  $ab^sc^t$ with $s+t=5$.
The inserted coefficients are the coefficients in the inequality to be shown. We can "dominate" with the positive coefficients on some "nodes" one (ans similarly more) negative coefficients on those nodes in the convex hull. For instance, a domination scheme can be applied by using from the positive positions marked with a bracket in
                b^6
                .
              .  .
            .   .  2
         [2][-1][-1][2]
        2  -1  -6  -1  .
      .   .  -1  -1  .  .
    .   .   .   2  2  .  . 
a^6                        c^6

and from each $[2]$ use only $[1]$ to dominate the $[-1]$ term.
Explicitly, we use:
$$
b^3(a^3-a^2c-ac^2+c^3)\ge 0\ .
$$
Use this pattern for all other $[-1]$ entries. Finally, the $-6$ in the middle is dominated by the remained positions. (Arithmetic mean is $\ge$ then geometric mean for instance.)

The claimed stronger inequality is false. Expanding, we would have to show equivalently an inequality corresponding to:
                b^6
                .
              .  .
            .  -1  1
          1   .   .  1
        1   .  -6   .  .
      .  -1   .   . -1  .
    .   .   .   1  1  .  . 
a^6                        c^6

But there is no chance to dominate the $-1$ entries. They escape from the convex hull of the positive entries. It is enough to consider the monomials in $a^4$. An inequality of the shape $a^4b^2-a^4bc\pm\dots\ge 0$ (where the dots cover $O(a^3)$) will never happen, just take $a=a(n)$ to be the polynomial $a(n)=n$, then $b,c$ constants with $c>b$, and pass with $n$ to infinity. Knowing how to produce the "bad case", makes it simple also to verify in the given inequality. The only terms having $a$ in numerator are in
$$
\frac ac-\frac{2a}{b+c}\ .
$$
Now use $a=a(n)=n$, $b=1$, $c=100$.
