Nessbit's and Triangle inequality A question given in a book was as follows:
Let $a$ $b$ and $c$ be the sides of a triangle. Prove that: 
$$\frac{3}{2}\leq \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}<2$$
My Attempt:
The LHS is the famous Nessbit's inequality whereas RHS is direct application of triangle inequality both of which i was able to do. My question is that:
Is there an inequality of the type $$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+\frac{r}{R}\leq 2,$$
where $r$ and $R$ being the inradius and circumradius of the triangle. 
If such an inequality exists then how to prove.
 A: Since
$$\frac{r}{R}=\frac{\frac{2S}{a+b+c}}{\frac{abc}{4S}}=\frac{16S^2}{2(a+b+c)abc}=\frac{(a+b+c)\prod\limits_{cyc}(a+b-c)}{2(a+b+c)abc}=\frac{\prod\limits_{cyc}(a+b-c)}{2abc},$$
we need to prove that
$$\sum_{cyc}\frac{a}{b+c}+\frac{\prod\limits_{cyc}(a+b-c)}{2abc}\leq2$$ or
$$\sum_{cyc}a^3(b+c)(a-b)(a-c)+\sum_{cyc}b^2c^2(b-c)^2\geq0,$$ which is true even for all non-negatives $a$, $b$ and $c$.
Indeed, let $a\geq b\geq c$.
Hence, $$\sum_{cyc}a^3(b+c)(a-b)(a-c)\geq a^3(b+c)(a-b)(a-c)+b^3(a+c)(b-a)(b-c)=$$
$$(a-b)(a^3(b+c)(a-c)-b^3(a+c)(b-c))=$$
$$=(a-b)(a^2(ab+ac)(a-c)-b^2(ab+bc)(b-c))\geq0.$$
Done!
The previous proof seems nice, but it bases on full expanding.
We can prove this inequality by SOS without full expanding.
Indeed, we need to prove that
$$\sum_{cyc}\frac{a}{b+c}+\frac{\prod\limits_{cyc}(a+b-c)}{2abc}\leq2$$ or
$$\sum_{cyc}\left(\frac{a}{b+c}-\frac{1}{2}\right)\leq\frac{1}{2}-\frac{\prod\limits_{cyc}(a+b-c)}{2abc}$$ or
$$\sum_{cyc}\frac{2a-b-c}{b+c}\leq\frac{\sum\limits_{cyc}(a^3-a^2b-a^2c+abc)}{abc}$$ or
$$\sum_{cyc}\frac{a-b-(c-a)}{b+c}\leq\frac{\sum\limits_{cyc}(2a^3-2a^2b-2a^2c+2abc)}{2abc}$$  or
$$\sum_{cyc}(a-b)\left(\frac{1}{b+c}-\frac{1}{a+c}\right)\leq\frac{\sum\limits_{cyc}(2a^3-2abc-2(a^2c+b^2c-2abc))}{2abc}$$ or
$$\sum_{cyc}\frac{(a-b)^2}{(a+c)(b+c)}\leq\frac{\sum\limits_{cyc}(a-b)^2(a+b+c-2c)}{2abc}$$ or
$$\sum_{cyc}(a-b)^2\left(\frac{(a+b-c)}{2abc}-\frac{1}{(a+c)(b+c)}\right)\geq0$$ or
$$\sum_{cyc}(a-b)^2(a+b)((a+b-c)(a+c)(b+c)-2abc)\geq0$$ or
$$\sum_{cyc}(a-b)^2(a+b)(a^2b+a^2c+b^2a+b^2c-abc-c^3)\geq0$$ or
$$\sum_{cyc}(a-b)^2(a+b)c(a^2+b^2-c^2)+\sum_{cyc}(a-b)^2(a+b)ab(a+b-c)\geq0.$$
Thus, it remains to prove that
$$\sum_{cyc}(a-b)^2(a+b)c(a^2+b^2-c^2)\geq0.$$
Now, let $a\geq b\geq c$.
Thus,
$$b^2\sum_{cyc}(a-b)^2(a+b)c(a^2+b^2-c^2)\geq$$
$$\geq b^2(a-c)^2(a+c)b(a^2+c^2-b^2)+b^2(b-c)^2(b+c)a(b^2+c^2-a^2)\geq$$
$$\geq a^2(b-c)^2(a+c)b(a^2-b^2)+b^2(b-c)^2(b+c)a(b^2-a^2)=$$
$$=ab(b-c)^2(a^2-b^2)(a(a+c)-b(b+c))\geq0.$$
Done again!
