# 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?

Let $$\{\alpha,\beta,\gamma\}=\{x,y,z\},$$ where $$x\geq y\geq z$$.
Thus, since $$\cos$$ decreases on$$(0,\pi)$$, by Rearrangement we obtain: $$\cos\alpha\sin\gamma+\cos\beta\sin\alpha+\cos\gamma\sin\beta\leq\cos{x}\sin{z}+\cos{y}\sin{y}+\cos{z}\sin{x}=$$ $$=\sin{(x+z)}+\frac{1}{2}\sin2y=\sin{y}+\frac{1}{2}\sin2y.$$ Can you end it now?
• @HY.fantasy I think your solution is not so good because you can not assume $A\geq B\geq C$. The inequality is cyclic and not symmetric. See please better my solution. Dec 16, 2019 at 18:29
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.