I've stumbled upon an exercise I don't really know how to handle. It's the following inequality:
If $A, B$ and $C$ are the angles of a triangle, prove $$\sin A \times \cos A + 2 \cos B \times \cos C \leq 1.$$
I've tried using triangle relations with its corresponding angles and other trigonometric formulas to reduce the problem to a simpler relation, but I've only made it worse so far.
Any help or suggestions? Cheers!
using the formulas $$S=\frac{bc}{2}\sin(A)$$ and $$\cos(A)=\frac{b^2+c^2-a^2}{2bc}$$ ect. you will have $$\frac{2A}{bc}\frac{b^2+c^2-a^2}{2bc}+2\frac{a^2+c^2-b^2}{2ac}\frac{a^2+b^2-c^2}{2ab}\le 1$$ and $$S=\sqrt{s(s-a)(s-b)(s-c)}$$ where $$s=\frac{a+b+c}{2}$$ we get $$4a^2(b^2+c^2-a^2)S\le bc(4a^2bc-2(a^2+c^2-b^2)(a^2+b^2-c^2))$$ now i would consider the cases: $$b^2+c^2=a^2$$ $$b^2+c^2<a^2$$ $$b^2+c^2>a^2$$ the first case gives us $$0\le (b-c)^2$$ which is true.