Trigonometric Inequality $\sin{A}+\sin{B}-\cos{C}\le\frac32$ To prove
$$\sin{A}+\sin{B}-\cos{C}\le\frac32$$
Given $A+B+C=\pi$ and $A,B,C>0$
I have managed to convert LHS to $$1-4\cos{\frac C2}\sin{\frac{A+C-B}2}\cos{\frac{A-B-C}2}$$ but that clearly isn't very helpful
One other conversion was $$\sin A+\sin B+\cos A\cos B-\sin A\sin B$$ but not sure how to proceed.
Any hints will be appreciated

Like In $ \triangle ABC$ show that $ 1 \lt \cos A + \cos B + \cos C \le \frac 32$ OR $ \cos {A} \cos {B} \cos {C} \leq \frac{1}{8} $
Let $y=\sin A+\sin B-\cos C=2\cos\dfrac C2\cos\dfrac{A-B}2-2\cos^2\dfrac C2+1$
$$\iff2\cos^2\dfrac C2-2\cos\dfrac C2\cos\dfrac{A-B}2+y-1=0$$
As $\cos\dfrac C2$ is real,the discriminant $\ge0$
$$4\cos^2\dfrac{A-B}2-8(y-1)\ge0\iff8y\le8+4\cos^2\dfrac{A-B}2\le8+4$$
