Using sin law we got equivalent form: $\frac{abc}{R^2} \leq (a+b+c)$ now using $[ABC]=\frac{abc}{4R}=\sqrt{S(S-a)(S-b)(S-c)}$ we arrive at: $8(S-a)(S-b)(S-c) \leq abc$ using substitution $x=b+c-a$, $y=c+a-b$, $z=a+b-c$ we have to prove: $xyz \leq \frac{1}{8} \cdot (x+y)(y+z)(z+x)$ which is obvious by AM-GM Equality holds when $a=b=c$