Because of simplicity let $A = a^2 + b^2 + c^2$ and $B = ab + bc + ca$. The CSB inequality gives us
$$\left( \frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} \right) (a(b+c) + b(c+a) + c(a+b)) \ge (a+b+c)^2,$$
i.e.
$$\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} \ge \frac{A+2B}{2B} = \frac{A}{2B} + 1.$$
Therefore,
$$\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} + \sqrt{\frac{ab+bc+ca}{a^2+b^2+c^2}} \ge \frac{A}{2B} + 1 + \sqrt{\frac{B}{A}}.$$
Finally, the inequality of arithmetic and geometric means implies
$$\frac{A}{2B} + \frac{1}{2}\sqrt{\frac{B}{A}} + \frac{1}{2}\sqrt{\frac{B}{A}} \ge 3\sqrt[3]{\frac{AB}{8BA}} = \frac{3}{2},$$
and the given inequality is proved.