An inequality with the sinus in a triangle.

Question

I have solved this problem in a way, rather "inspired". I would like to have a solution found an easier way but I was unable so far.

Let $A,B,C$ the angles of a triangle $\triangle {ABC}$; prove that $$(\sin^2 A+\sin^2 B+\sin^2 C)\le \frac14 \left(\frac{1}{\sin A}+\frac{1}{\sin B}+\frac{1}{\sin C}\right) (\sin A+\sin B+\sin C)$$

Answer 1

Wee need to prove that $$\sum\limits_{cyc}\frac{4S^2}{b^2c^2}\leq\frac{1}{4}\sum\limits_{cyc}\frac{bc}{2S}\sum\limits_{cyc}\frac{2S}{bc}$$ or $$\frac{(a^2+b^2+c^2)\sum\limits_{cyc}(2a^2b^2-a^4)}{a^2b^2c^2}\leq\frac{(ab+ac+bc)(a+b+c)}{abc}$$ or $$abc(ab+ac+bc)\geq(a^2+b^2+c^2)(a+b-c)(a+c-b)(b+c-a)$$ or $$\sum\limits_{cyc}(a^5-a^4b-a^4c+2a^3bc-a^2b^2c)\geq0,$$ which is true even for all non-negatives $a$, $b$ and $c$.

Indeed, $\sum\limits_{cyc}(a^5-a^4b-a^4c+a^3bc)\geq0$ is true by Schur and

$\sum\limits_{cyc}(a^3bc-a^2b^2c)=\frac{1}{2}abc\sum\limits_{cyc}(a-b)^2\geq0$.

Done!