Maximum value of $\sin (x)\;\sin (2x)\;\sin (3x)$ To prove : $\sin (x)\;\sin (2x)\;\sin (3x) \lt 9/16$
With some transformations i was able to prove the above result using calculus but I am not getting the way to solve it without the use of calculus
I've tried grouping $\sin(x)\;\sin(3x)$ and then using product to sum transformations.
Also I've tried to find the maximum value of sinxsin3x and then using the fact that sin2x is a fraction. Although I was able to prove that sinxsin3x is less than 9/16 but the minimum value was less than -9/16 and since sin2x can be negative so i could not draw any conclusion.
 A: Let $a= \sin x ,b = \cos x$ and $a^2+b^2=1$,
Since $\sin 2x = 2 \sin x \cos x = 2 a b$
And $\sin 3x = \sin x (2\cos x+1)(2\cos x-1) = a (2b+1)(2b-1)$
Thus we have $\sin x \sin 2x \sin 3x = 8a^3 b^3 -2a^3 b$
Substitute instead of $b = +\sqrt{1-a^2}$ and once $b=-\sqrt{1-a^2}$
Solving the inequality $8a^3 b^3 -2a^3 b < \frac{9}{16}$ will be easy from here.
A: Let $\cos^2x=t$.
Thus, we need to prove that
$$2\sin^3x\cos{x}(3-4\sin^2x)<\frac{9}{16}$$ or
$$\sin^3x\cos{x}(4\cos^2x-1)<\frac{9}{32},$$
which is obviously true for
$$\sin^3x\cos{x}(4\cos^2x-1)\leq0.$$
Now, let $\sin^3x\cos{x}(4\cos^2x-1)>0$.
Thus, it's enough to prove that
$$t(1-t)^3(4t-1)^2<\frac{81}{1024}.$$
1. Let $\frac{1}{4}<t<1.$
Thus, we need to prove that $f(t)<\ln\frac{81}{1024},$ where
$$f(t)=\ln{t}+3\ln(1-t)+2\ln(4t-1).$$
Indeed, $$f'(t)=\frac{1}{t}-\frac{3}{1-t}+\frac{8}{4t-1}=-\frac{24t^2-16t+1}{(1-t)t(4t-1)},$$
which gives $t_{max}=\frac{4+\sqrt{10}}{12}$ and since $f\left(\frac{4+\sqrt{10}}{12}\right)<\ln\frac{81}{1024},$ we are done in this case;


*$0<t<\frac{1}{4}$.


The work in this case is the same: $g(t)=\ln{t}+3\ln(1-t)+2\ln(1-4t)$...
