<s>[INST] Show that $\sin (x/2)+2\sin (x/4)\leq3\sqrt{3}/2$ for all $x \in [0,2\pi]$ 
Given that  $$f (x)=\sin (x/2)+2\sin (x/4)$$
  Show that $f (x)\leq3\sqrt{3}/2$   for all $x \in [0,2\pi]$.

My try 
$f (x)=\sin (x/2)+2\sin (x/4)$
$f' (x)=\frac {1}{2}\cos(x/2)+\frac {1}{2}\cos (x/4)$
$f' (x)=\frac {1}{2}\cos(x/2)+\frac {1}{2}\cos (x/4)$
$f' (x)=\frac {1}{2}(\cos(x/2)+\cos (x/4))$
$f' (x)=\cos(3x/4)\cos (x/4))$
$f'(x)=0$   $\implies$  $\cos(3x/4)\cos (x/4))=0$
$\cos(3x/4)=0\:$   or $\:\cos (x/4)=0$
$3x/4= \pi/2\:$  and so $\:x= 2\pi/3$      or $ x=2\pi$
$x/4= \pi/2\:$  and so $\:x= 2\pi $
Is my work ok ? What should I do now? 
 [/INST] Hint:
Set $x=8y$  to find $$2\sin2y(1+\cos2y)=8\sin y\cos^3y$$
Now $\sqrt[4]{3\sin^2y\cos^6y}\le\dfrac{3\sin^2y+\cos^2y+\cos^2y+\cos^2y}4=\dfrac34$
Squaring we get  $$16\sqrt3\sin y\cos^3y\le9$$
The equality occurs if $3\sin^2y=\cos^2y\iff\dfrac{\sin^2y}1=\dfrac{\cos^2y}3=\dfrac1{1+3}$
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