In a $\triangle ABC,\angle B=60^{0}\;,$ Then range of $\sin A\sin C$ 
In a $\triangle ABC,\angle B=60^{0}\;,$ Then range of $\sin A\sin C$ 

$\bf{My\; Attempt:}$
Using Sin formula:
$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$
So $\displaystyle \frac{a}{\sin A} = \frac{b}{\sin 60^0}\Rightarrow \sin A = \frac{a}{b}\cdot \frac{\sqrt{3}}{2}$ and $\displaystyle \frac{c}{\sin C} = \frac{b}{\sin 60^0}\Rightarrow \sin C = \frac{c}{b}\cdot \frac{\sqrt{3}}{2}$
So $$\sin A\sin C = \frac{3}{4}\cdot \frac{ac}{b^2}$$
Now using Cosine formula:
$$\cos B  = \cos 60^0 = \frac{1}{2}= \frac{a^2+c^2-b^2}{2ac}\Rightarrow b^2=a^2+c^2-ac$$
So $$\sin A\sin C = \frac{3}{4}\bigg[\frac{ac}{a^2+c^2-ac}\bigg] = \frac{3}{4}\bigg[\frac{1}{\frac{a}{c}+\frac{c}{a}-1}\bigg]\leq \frac{3}{4}$$
Using $\bf{A.M\geq G.M},$ We get $$\frac{a}{c}+\frac{c}{a}\geq 2\Rightarrow \frac{a}{c}+\frac{c}{a}-1\geq 1$$
$\bf{ADDED::}$ Using Jensen Inequality:: For $f(x) = \ln(\sin x)\;,$
$$\ln(\sin A)+\ln(\sin C)\leq 2\cdot \ln \sin \left(\frac{A+C}{2}\right)  = \ln \cos 30^0 = 2\cdot \ln \frac{\sqrt{3}}{2} = \ln \frac{3}{4}$$
But I do not understand how to calculate the lower bound for $\sin A\sin C$.
Thanks in advance!

The range of the function is $0 < f \le 3/4$.
You already figured out the maximum value of $f$ so I will not show the solution for that part.
As angle $A$ gets closer and closer to $0$ degrees, the value of $\sin(A)$ tends to $0$. $\sin(C)$ will tend towards a finite constant, since $C$ is going towards $120$ degrees. Thus, the product will tend to $0$. $\lim_{\theta -> 0}{f = 0}$. The function has no exact minimum value because it can never reach $0$, but it will get infinitely close to $0$.
