In triangle $ABC$, find maximum value of $\sin A \cos B + \sin B \cos C + \sin C \cos A$ 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? 

You can use Lagrange multipliers. You are maximizing 
$$
\sin A\cos B+\sin B\cos C+\sin C\cos A
$$
under
$$
A+B+C=\pi.
$$
Using Lagrange multipliers and the identity $\cos (x+y)=\cos x\cos y-\sin x\sin y$, you easily get
$$
\cos(A+B)-\lambda=\cos(B+C)-\lambda=\cos(A+C)-\lambda=0,
$$
from where it follows that $\cos C=\cos A=\cos B=-\lambda$ and then necessarily (given that at least two of them are acute) $A=B=C$ and the maximum is achieved for the equilateral triangle. 
