In fact, if I am not mistaken in my calculations, for this M we have $\mathrm{MA}\times \mathrm{MB}\times \mathrm{MC}=8{\sin }^{2}A/2\cos (B-C)/2=4\sin A/2(\cos B+\cos C)$.

Since $A>B\ge C$, we have $A>\pi /3$, so $\sin A/2>1/2$; thus it suffices to show that $\cos B+\cos C>1$.

This looks like it ought to be easy, but I haven't found a sensible proof. If we use the formula for cosine in terms of the side lengths we must show

$b(a^2+c^2-b^2)+c(a^2+b^2-c^2)>2abc$

Expanding everything out we must show

$a^{2}b+b{c}^2+a^{2}c+b^{2}c>2abc+b^3+c^3$

We know that $a^{2}b>b^3$ and $b{c}^2>c^3$ so it suffices to show that

$a^{2}c+b^{2}c>2abc$

which is just $c(a-b{)}^2>0$ and is consequently true.

So we are done. (Phew!)

I'll leave you the fun of checking that MA * MB * MC is never > 2 for an equilateral triangle (use Ptolemy's theorem).