Denote $x=\frac{b+c-a}{2}>0,...$, that is perform Ravi substitution. Our condition is $(x+y)(y+z)(z+x)=8$ and the inequality $$\sum_{cyc}\sqrt{\frac{x(x+y)(z+x)}{y+z}}\ge\sqrt2\cdot\sum_{cyc}x.$$Since $x,y,z\in\mathbb{R}_+$ there holds $$\sum_{cyc} xy(x-y)^2\ge0\iff$$$$(x+y+z)(x+y)(y+z)(z+x)\ge2\left(x^2(y+z)^2+y^2(x+z)^2+z^2(x+y)^2\right)\iff$$$$x+y+z\ge 2\sum_{cyc}\frac{x^2(y+z)}{(x+y)(z+x)}>0.$$Holder's inequality yields $$\left(\sum_{cyc}\sqrt{\frac{x(x+y)(z+x)}{y+z}}\right)^2\cdot\sum_{cyc}\frac{x^2(y+z)}{(x+y)(z+x)}\ge\left(\sum_{cyc}x\right)^3>0.$$Multiply two last inequalities and take a square root.$\blacksquare$