Maximizing $\sin \beta \cos \beta + \sin \alpha \cos \alpha - \sin \alpha \sin \beta$ I need to maximize
$$ \sin \beta \cos \beta + \sin \alpha \cos \alpha - \sin \alpha \sin \beta \tag{1}$$
where $\alpha, \beta \in [0, \frac{\pi}{2}]$.
With numerical methods I have found that
$$ \sin \beta \cos \beta + \sin \alpha \cos \alpha - \sin \alpha \sin \beta \leq 2 \sin \frac{\alpha + \beta}{2} \cos \frac{\alpha + \beta}{2} - \sin^2 \frac{\alpha + \beta}{2}. \tag{2} $$
If $(2)$ is true then I can denote $x = \frac{\alpha + \beta}{2}$ and prove (using Cauchy inequality) that 
$$ 2 \sin x \cos x - \sin^2 x \leq \frac{\sqrt{5}-1}{2}. \tag{3}$$
But is $(2)$ true? How do I prove it? Maybe I need to use a different idea to maximize $(1)$?
 A: Let $$f = \sin \alpha \cos \alpha + \sin \beta \cos \beta -\sin \alpha \sin \beta = \frac{1}{2}(\sin 2\alpha + \sin 2\beta) -\sin \alpha \sin \beta$$
$$\frac{\partial f}{\partial \alpha} = 0 \implies \cos 2\alpha - \sin \beta \cos \alpha = 0$$
$$\frac{\partial f}{\partial \beta} = 0 \implies \cos 2\beta - \sin \alpha \cos \beta = 0$$
From symmetry (also may not be difficult to show directly) $$\beta = \alpha$$
Finally, $$\alpha = \beta = \frac{1}{2} \tan^{-1} 2$$
Substituting in $f$, the maximum value is 0.61803398874989484820
A: As Dan suggested, solve for the stationary points of the gradient. You'll end up solving two (identical) quartics in $\sin \alpha$, $\sin \beta$ (which are really just 2 quadratics due to the symmetry) and find that the only stationary point in the range is at $\alpha = \beta = \pi/6$. For this to not be the maximum on the domain, the max must occur on the boundary, which are much simpler sub-problems to solve (and by the symmetry you only have to solve 2), and in fact will prove that $\alpha = \beta = \pi/6$ gives the max over the domain.
A: A nice solution with elementary methods was provided by user arqady in AoPS (see here). It is as follows:
It is easy to see, using the trigonometric addition formulas that
$$ \sin \beta \cos \beta + \sin \alpha \cos \alpha - \sin \alpha \sin \beta=\sin(\alpha+\beta)\cos(\alpha-\beta)-\frac{1}{2}\cos(\alpha-\beta)+\frac{1}{2}\cos(\alpha+\beta) $$
Using Cauchy's inequality,
$$ \sin(\alpha+\beta)\cos(\alpha-\beta)-\frac{1}{2}\cos(\alpha-\beta)+\frac{1}{2}\cos(\alpha+\beta) \leq \sqrt{\cos(\alpha - \beta)^2+\frac{1}{4}}-\frac{1}{2} \cos (\alpha - \beta) $$
and, denoting $\cos(\alpha - \beta) = x$ 
$$ \sqrt{x^2+\frac{1}{4}}-\frac{1}{2} x \leq\frac{\sqrt5-1}{2} $$
because, after squaring,
$$3x^2 +2x - 2\sqrt{5}x + 2\sqrt{5} - 5 \leq 0 \\
(x-1)(3x+5-2\sqrt{5}) \leq 0$$
which holds because $\alpha, \beta \in [0, \frac{\pi}{2}]$ so $x \in [0, 1]$. (The first term is nonnegative, the second one is positive.)
Equality holds if $\frac{\sin(\alpha + \beta)}{\cos(\alpha - \beta)} = \frac{\cos(\alpha + \beta)}{\frac{1}{2}}$ and $x = 1$. Therefore $\alpha = \beta$ and $\tan 2\alpha = 2$ so 
$$ \sin \beta \cos \beta + \sin \alpha \cos \alpha - \sin \alpha \sin \beta \leq  \frac{\sqrt5-1}{2}$$
with equality at $\alpha = \beta = \frac{1}{2} \arctan 2$.
