Solve the maximum of a trigonometric function without differentiation

Question

The problem is, when $f(x) \stackrel{\mathrm{def}}{=} \frac{\sin(x)\cos(x)}{(1+\sin(x))^2} $for $(x \in [0,\frac π2])$,

what is the maximum of $f(x)$？

Of course, we solve this problem with differentiation. Specifically, to solve $f’(x)=0$. The answer is $sqrt(3)/9$ when $x=π/6$

I am also searching for a solution using inequalities(e.g. AM-GM, Cauchy-Schwarz).

However, I can't having precisely applying inequalities. This is Because, I think, $f(x)$ is not symmetric.

Would you mind to tell me how to solve this problem using inequality?

Answer 1

By AM-GM, we have $$\left(\frac{\sin x\, \cos x}{(1+\sin x)^2}\right)^2 = \frac{\sin x\cdot \sin x\cdot (1 - \sin x)}{(1 + \sin x)^3} \le \frac{\frac{1}{27}[\sin x + \sin x + (1 - \sin x)]^3}{(1 + \sin x)^3} = \frac{1}{27}$$ which results in $$\frac{\sin x\, \cos x}{(1+\sin x)^2}\le \frac{\sqrt{3}}{9}.$$

Answer 2

The most simple representation is obtained observing that the denominator is a form of the $1+\cos(x)$, so translate by $-\pi/2$ and double the angle.

$$\frac{\sin \left(2 \left(2 x+\frac{\pi }{2}\right)\right)}{\left(\sin \left(2 x+\frac{\pi }{2}\right)+1\right)^2}=\tan ^3(x)-\tan (x)$$

But the function is not even by reflection at its maximum

$$(x\to x^3-x)[(x\to \tan x^3)[x]] $$

$$\tan \left(-\frac{\pi }{6}\right)=\frac{1}{\sqrt{3}}$$

$$\left(y \to y^3-y\right)\left(\frac{1}{\sqrt{3}}\right)=\frac{2}{3 \sqrt{3}}$$