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!
 A: Please check the following hint. Firstly we have 
$$
R := \sin A \sin C = \sin A \sin \left( \pi - (B+A) \right) =  \sin A \sin \left( \frac{2\pi}{3} - A \right)
$$
thus, 
\begin{align}
R 
&= \sin A \left( \frac{\sqrt{3}}{2} \cos A + \frac{1}{2} \sin A \right)
= \frac{\sqrt{3}}{4} \sin 2A + \frac{1}{4} \left( 1 - \cos 2A \right)
\\
&= 
\frac{\sqrt{3}}{4} \sin 2A - \frac{1}{4} \cos 2A + \frac{1}{4}
= 
\frac{1}{2}\left( \sin 2A \cos \frac{\pi}{6} - \cos 2A \sin \frac{\pi}{6} \right) + \frac{1}{4}
\\
&
= 
\frac{1}{2}\sin \left( 2A - \frac{\pi}{6} \right) + \frac{1}{4}
\end{align}
Besides, 
$$
C  = \pi - (A+B) = \frac{2\pi}{3} - A > 0 
$$
hence, $0 <A < \frac{2\pi}{3}$, and 
$$
0 < 2A - \frac{\pi }{6} < \frac{4\pi}{3} - \frac{\pi }{6} = \pi + \frac{\pi }{6}
$$
Then
$$
- \frac{1}{2} < \sin \left( 2A - \frac{\pi}{6} \right) \le 1
$$
Lastly we obtain
$$
0 < R = \frac{1}{2} \sin \left( 2A - \frac{\pi}{6} \right) + \frac{1}{4}\le \frac{3}{4}
$$
In addition, $R = 3/4$ when $2 A - \frac{\pi}{6} = \frac{\pi}{2}$ or $A = \frac{\pi}{3}$ (i.e. $A=B=C=\pi/3$).
EDIT: There was a mistake right here: $0 <A < \frac{2\pi}{3}$, and 
$$
- \frac{\pi }{6}  < 2A - \frac{\pi }{6} < \frac{4\pi}{3} - \frac{\pi }{6} = \pi + \frac{\pi }{6}
$$
However, in this case, we also have 
$$
- \frac{1}{2} < \sin \left( 2A - \frac{\pi}{6} \right) \le 1
$$
Then the remainder of solution is unchanged.
A: Clearly $\sin(A)\sin(C)\geq 0$ since $A$ and $C$ are between $0^{\circ}$ and $180^{\circ}$. Let $A$ approach $0^{\circ}$. Then $\sin(A)$ approaches $0$ as well, while $\sin(C)$ is bounded above by $1$. This shows that
$$\sin(A)\sin(C)\rightarrow 0$$
if $A\rightarrow 0^{\circ}$ so that $\sin(A)\sin(C)$ can be as small as needed. The lower bound is $0$.
A: 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 $0$.
A: It is sufficient to consider the function
$$f(x)=\sin (x)\sin(\frac{2\pi}{3}-x)$$ restreint to the domain $D=\{x|\space 0\lt x\lt\dfrac{2\pi}{3}\}$. 
It is easy to find $f$ is positive and has a maximun at the point $x=\dfrac{\pi}{3}\in D$; furthermore the infimum of $f(x)$ is equal to $0$ taken to the neighborhood of both  $0$ and $\dfrac{2\pi}{3}$.
Thus the range of $f$ is the semi-open interval $(0,\space\frac34]$.
