In triangle $ABC$, find maximum value of $$\sin A \cos B + \sin B \cos C + \sin C \cos A$$
We could make $\cos C = - \cos(A+B)$ and $\sin C = \sin(A+B)$.
But then we have a rather awkward expression that doesn't share the same power
$$ \sin A \cos B + \sin A \sin^{2}B + \sin B \cos^2A $$
The answer btw is not hard to be guessed, $\frac{3}{4} \sqrt{3}$, but not sure how to prove it.
Actually looks like we can solve like below:
if $A < B < C$, then $cosA > cosB > cosC$, $sinC > sinB > sinA$
Then $sinA cosB + sinB cosC + sinC cosA \leq cosAsinC + sinBcosB + sinAcosC = sinBcosB + sinB = sinB \sqrt{1-sin^2 B} + sinB$
did I do this correctly?