Showing that the maximum value of $\sin x+\sin y\sin z$, where $x+y+z=\pi$, is the golden ratio 
Find the maximum of
$$\sin x+\sin y\sin z$$
if  $x+y+z=\pi$.

By using Lagrange multipliers, concluded that $y=z$, further plug $x=\pi-2y$, I've reduced the problem to single-variable expression.
$$\sin^2 y+\sin(2y)$$
Then by taking first derivative and using formulas for $\sin(\arctan x)$ and $\cos(\arctan x)$, finally we can obtain the maximum
$$\frac{\sqrt5+1}{2}$$
As one can see this is precisely the golden ratio $\phi$!
But this solution takes some time, so I'm interested in different no calculus solution of this problem, especially considering that it is related to the golden ratio.
 A: Using algebra and trigonometry only:
First observe that since the angles sum to $\pi$ radians, $\sin x=\sin(y+z)$.  Then, apply the product-sum relation
$\sin y\sin z=(1/2)[\cos(y-z)-\cos(y+z)]$
Then the objective function may then be rendered as
$\sin x+\sin y\sin z =\color{blue}{\sin(y+z)-(1/2)\cos(y+z)}+\color{brown}{(1/2)\cos(y-z)}$
The blue expression has the form $a\sin\theta-b\cos\theta$ for positive $a$ and $b$, therefore $\sqrt{a^2+b^2}\sin[\theta-\arccos(a/\sqrt{a^2+b^2})]$.  Thus the blue expression is maximized at $\sqrt5/2$ whenever $y+z=\pi/2+\arccos(2/\sqrt5)=\pi-\arcsin(2/\sqrt5)$.  If we then select each of $y$ and $z$ as half of this sum, then the brown term is simultaneously maximized at $1/2$.  Thereby a maximum value of $(\sqrt5+1)/2$ is achieved.
A: Substituting
$z=-x-y+\pi$
into the first equation, we get:
$\frac{\sqrt{5}}{2}sin(atan(\frac{1}{2})+x)-\frac{cos(x+2y)}{2}$
Setting
$atan( \frac{1}{2})+x=\frac{\pi}{2}$
and
$x+2y=\pi$
to make maximum both $sin$ and $cos$, we get:
$x=atan(\frac{1}{3})+\frac{\pi}{4}$
and
$y=\frac{3\pi}{8}-\frac{atan(\frac{1}{3})}{2}$
that substituted in the initial function , is as a result:
$\frac{\sqrt{5}+1}{2}$,
the golden ratio.
