Prove that in any triangle 
Prove that in any triangle with side lengths $a,b,c$ inradius $r$, and
  circumradius $R$, we have
$\frac{a}{b+c}+ \frac{b}{c+a}+ \frac{c}{a+b}+ \frac{r}{R} >
\frac{5}{3}$

I think it is $\frac{a}{b+c}+ \frac{b}{c+a}+ \frac{c}{a+b}+ \frac{r}{R} \leq 2 $.
 with $a+b+c=2s,\ ab+b+ca=(4R+r)r+s^2,\ abc=4Rrs$ rewrites as$$\frac{2\left(s^2-Rr-r^2\right)}{s^2+2Rr+r^2}+\frac{r}{R}-\frac{5}{3}\ge 0$$which boils down to showing$$(R+3r)s^2-\left(16R^2+5Rr-3r^2\right)r\ge 0$$which by Gerretsen $s^2\ge 16Rr-5r^2$ is just$$2(19R-6r)r^2\ge 0$$true by Euler $R\ge 2r$.

In the standard notation we need to prove that
$$\sum_{cyc}\frac{a}{b+c}+\frac{\frac{2S}{a+b+c}}{\frac{abc}{4S}}\leq2$$ or
$$\sum_{cyc}\frac{a}{b+c}+\frac{16S^2}{2abc(a+b+c)}\leq2$$ or
$$\sum_{cyc}\frac{a}{b+c}+\frac{\sum\limits_{cyc}\left(-a^3+a^2b+a^2c-\frac{2}{3}abc\right)}{2abc}\leq2$$ or
$$\sum_{cyc}(a^5b+a^5c-2a^3b^3-2a^4bc+a^3b^2c+a^3c^2b)\geq0$$ or
$$\sum_{cyc}ab(a^2-b^2)^2-abc\sum_{cyc}(a+b)(a-b)^2\geq0$$ or
$$\sum_{cyc}ab(a+b)(a-b)^2(a+b-c)\geq0$$ and we are done!
