hello, can you correct your $\LaTeX$ formula please. Sonnhard.

Let $[ABC]=K$. It is easy to prove that $r_ar_br_c=sK$ using the formulae $K=(s-a)r_a=(s-b)r_b=(s-c)r_c$ and $K=\sqrt{s(s-a)(s-b)(s-c)}$, where $r_a,r_b,r_c$ are the exradii of $A,B,C$ respectively. So we aim to prove $sK\le \frac{3\sqrt{3}}{8}abc$. Using the formula $K=\frac{abc}{4R}$ it becomes $\frac{a+b+c}{2R}\le 3\sqrt{3}$. now from the sine rule $\frac{a+b+c}{2R}=\sin A + \sin B + \sin C$, and Jensen finishes it off.

Let $r_a,r_b,r_c$ denotes the radius of the excribed circles corresponding to the sides of lengths $a, b, c$ respectively. If P be the semiperimeter of the triangle, then $\sqrt{P(P-a)(P-b)(P-c)}= \frac{abc}{4R}=r_a(P-a)=r_b(P-b)=r_c(P-c)$. So the well known inequality $r_ar_br_c\le \frac{3\sqrt{3}}{8}abc$ transforms into $P\le \frac{3\sqrt{3}{R}}{2}$ which is equivalent to $sin\alpha + sin\beta + sin\gamma \le \frac{3\sqrt{3}}{2}$ which is just the Jensen inequality for $sin(x)$ in a triangle.