I proved this inequality in this way $$\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{b+c}{2}\left(\frac{1}{c+a}+ \frac{1}{a+b}\right)\geq\frac{b+c}{\sqrt{(c+a)(a+b)}},\cdots$$ $$ \left(\frac{b}{c+a}+\frac{c}{a+b}\right)\left( \frac{c}{a+b}+\frac{a}{b+c}\right)\left( \frac{a}{b+c}+\frac{b}{c+a}\right)\ge1$$ $$\sum \sqrt[3]{\dfrac{a}{b+c}+\dfrac{b}{c+a}}\geq 3 \sqrt[9]{ \left(\frac{b}{c+a}+\frac{c}{a+b}\right)\left( \frac{c}{a+b}+\frac{a}{b+c}\right)\left( \frac{a}{b+c}+\frac{b}{c+a}\right)}\geq 3$$
But i found this inequality proved in his way here
Note
$$ \left( {\frac {b}{c+a}}+{\frac {c}{a+b}} \right) \left( b \left( a+b \right) +c \left( c+a \right) \right) ^{3} \left( c+a \right) \left( a+b \right) = \left( ab+{b}^{2}+{c}^{2}+ac \right) ^{4} $$ By Holder,we have
$$\left( {\it \sum} \left( \sqrt [3]{{\frac {b}{c+a}}+{\frac {c}{a+b}}} \right) \right) ^{3}{\it \sum} \left( \left( ab+{b}^{2}+{c}^{2}+ac \right) ^{3} \left( c+a \right) \left( a+b \right) \right) \geq \left( {\it \sum} \left( ab+{b}^{2}+{c}^{2}+ac \right) \right) ^{4}$$
Need to prove
$${\frac { \left( {\it \sum} \left( ab+{b}^{2}+{c}^{2}+ac \right) \right) ^{4}}{{\it \sum} \left( \left( ab+{b}^{2}+{c}^{2}+ac \right) ^{3} \left( c+a \right) \left( a+b \right) \right) }}\geq 27$$ <=>
$ \left( {\it \sum} \left( ab+{b}^{2}+{c}^{2}+ac \right) \right) ^{4}\geq 27\,{\it \sum} \left( \left( ab+{b}^{2}+{c}^{2}+ac \right) ^{3} \left( c+a \right) \left( a+b \right) \right) $
<=>(In $\triangle ABC$,subs...)
$a->s-a,b->s-b,c->s-c$ <=> $$f=216\,{s}^{8}-1296\,r \left( r+2\,R \right) {s}^{6}+1728\,{r}^{2} \left( 4\,Rr+2\,{R}^{2}+{r}^{2} \right) {s}^{4}-48\,{r}^{3} \left( 4 \,R+r \right) \left( 13\,{r}^{2}+158\,Rr+136\,{R}^{2} \right) {s}^{2} +232\,{r}^{4} \left( 4\,R+r \right) ^{4}\geq 0$$
But $$ f={{\it u2}}^{2} \left( 648\,{r}^{2}{s}^{2}+864\,r{s}^{2}R \right) +{ \it u1}\,{\it u2}\, \left( 149232\,{r}^{5}+64704\,R{r}^{4}+56832\,{R}^ {2}{r}^{3}+6912\,{s}^{2}{r}^{3} \right) +{\it u1}\, \left( 130624\,{r} ^{7}+968704\,{R}^{3}{r}^{4}+115008\,{R}^{2}{r}^{5} \right) +{\it u2}\, \left( 216\,{s}^{6}+19008\,{s}^{2}{R}^{2}{r}^{2}+350760\,{r}^{6} \right) +{\it u3}\, \left( 203568\,R{r}^{5}+3024\,{r}^{2}{s}^{4} \right) \geq 0$$
as ${\it u1}=R-2\,r,{\it u2}={s}^{2}-16\,Rr+5\,{r}^{2},{\it u3}=4\,{R}^{2 }+4\,Rr+3\,{r}^{2}-{s}^{2} $ Done!
Here i did not understood how did he sustituted in this portion
<=>(In $\triangle ABC$,subs...) $a->s-a,b->s-b,c->s-c$\ <=>$$f=216\,{s}^{8}-1296\,r \left( r+2\,R \right) {s}^{6}+1728\,{r}^{2} \left( 4\,Rr+2\,{R}^{2}+{r}^{2} \right) {s}^{4}-48\,{r}^{3} \left( 4 \,R+r \right) \left( 13\,{r}^{2}+158\,Rr+136\,{R}^{2} \right) {s}^{2} +232\,{r}^{4} \left( 4\,R+r \right) ^{4}\geq 0$$But$$f={{\it u2}}^{2} \left( 648\,{r}^{2}{s}^{2}+864\,r{s}^{2}R \right) +{ \it u1}\,{\it u2}\, \left( 149232\,{r}^{5}+64704\,R{r}^{4}+56832\,{R}^ {2}{r}^{3}+6912\,{s}^{2}{r}^{3} \right) +{\it u1}\, \left( 130624\,{r} ^{7}+968704\,{R}^{3}{r}^{4}+115008\,{R}^{2}{r}^{5} \right) +{\it u2}\, \left( 216\,{s}^{6}+19008\,{s}^{2}{R}^{2}{r}^{2}+350760\,{r}^{6} \right) +{\it u3}\, \left( 203568\,R{r}^{5}+3024\,{r}^{2}{s}^{4}\right) \geq 0$$as$${\it u1}=R-2\,r,{\it u2}={s}^{2}-16\,Rr+5\,{r}^{2},{\it u3}=4\,{R}^{2 }+4\,Rr+3\,{r}^{2}-{s}^{2}$$Done!
Let $z = a + b$, $y = c+a$ and $x = b+c$ (correspondingly, $a = \frac{y+z-x}{2}, b = \frac{z+x-y}{2}$ and $c = \frac{x+y-z}{2}$). Then $x, y, z$ are side lengths of a triangle.
We need to prove that $f(a, b, c) \ge 0$ for $a, b, c\ge 0$. Equivalently, we turn to prove that $f(x, y, z)\ge 0$ where $x, y, z$ are side lengths of a triangle.
We have the following relation: \begin{align} &\left\{\begin{array}{l} s = \frac{x+y+z}{2} \\[4pt] R = \frac{xyz}{\sqrt{(x+y+z)(x+y-z)(y+z-x)(z+x-y)}} \\[6pt] r = \frac{\sqrt{(x+y+z)(x+y-z)(y+z-x)(z+x-y)}}{2(x+y+z)} \end{array} \right.\\[10pt] &\Longleftrightarrow \left\{\begin{array}{l} x+y+z = 2s, \\ xy+yz+ zx = s^2 + 4Rr + r^2 \\ xyz = 4sRr. \end{array} \right. \end{align} There are many relations among $s, R, r$, e.g., $R \ge 2r$ (Euler's inequality).
Then, $f(x, y, z)$ is written in terms of $s, R, r$. Now we need to prove that $f(s, R, r)\ge 0$.