$\sqrt{(a+b-c)(b+c-a)(c+a-b)} \le \frac{3\sqrt{3}abc}{(a+b+c)\sqrt{a+b+c}}$ Suppose $a, b, c$ are the  lengths of three triangular edges. Prove that:
$$\sqrt{(a+b-c)(b+c-a)(c+a-b)} \le \frac{3\sqrt{3}abc}{(a+b+c)\sqrt{a+b+c}}$$

As the hint give in the comment says (I denote by $S$ the area of $ABC$ and by $R$ the radius of its circumcircle), if you multiply your inequality by $\sqrt{a+b+c}$ you'll get
$$4S \leq \frac{3\sqrt{3}abc}{a+b+c}$$
which is eqivalent to
$$a+b+c \leq 3\sqrt{3}\frac{abc}{4S}=3\sqrt{3}R.$$

This inequality is quite known. If you want a proof, you can write $a=2R \sin A$ (and the other two equalities) and get the equivalent inequality
$$ \sin A +\sin B +\sin C \leq \frac{3\sqrt{3}}{2}$$
which is an easy application of the Jensen inequality for the concave function $\sin : [0,\pi] \to [0,1]$.
