In a triangle prove that $\mathrm{cosec}( A) +\mathrm{cosec}(B) +\mathrm{cosec}(C)\le \frac{2\sqrt 3}{9}\left (1+\frac Rr\right)^2$ In a triangle prove that $\mathrm{cosec}( A) +\mathrm{cosec}(B) +\mathrm{cosec}(C)\le \frac{2\sqrt 3}{9}\left (1+\frac Rr\right)^2$
My progress:
$\left (1+\frac Rr\right)^2=\left(\frac{2\prod sin A+\sum sin A}{2\prod sin A}\right)^2\ge \left(\frac{2\prod sin A+\sum sin A sin B}{2\prod sin A}\right)^2\ge\frac{8\prod sin A\sum sin A sin B} {\left(2\prod sin A\right)^2}=2\sum cosec A$
 A: Let $a=\frac{y+z}{2},$ $b=\frac{x+z}{2}$ and $c=\frac{x+y}{2}$
Thus, in the standard notation  we need to prove that:
$$\sum_{cyc}\frac{1}{\sin\alpha}\leq\frac{2}{3\sqrt3}\left(1+\frac{\frac{abc}{4S}}{\frac{2S}{a+b+c}}\right)^2$$ or
$$\sum_{cyc}\frac{1}{\frac{2S}{bc}}\leq\frac{2}{3\sqrt3}\left(1+\frac{2abc(a+b+c)}{16S^2}\right)^2$$ or
$$\frac{ab+ac+bc}{4S}\leq\frac{1}{3\sqrt3}\left(1+\frac{2abc}{\prod\limits_{cyc}(a+b-c)}\right)^2$$ or
$$\frac{\sum\limits_{cyc}(x+y)(x+z)}{4\sqrt{xyz(x+y+z)}}\leq\frac{1}{3\sqrt3}\left(1+\frac{\prod\limits_{cyc}(x+y)}{4xyz}\right)^2$$ or
$$12xyz\sum\limits_{cyc}(x^2+3xy)\cdot\sqrt{\frac{3xyz}{x+y+z}}\leq\left(\sum_{cyc}(x^2y+x^2z+2xyz)\right)^2.$$
But $$\sum_{cyc}(x^2+3xy)\leq\frac{4}{3}(x+y+z)^2$$ and $$\sqrt{3xyz(x+y+z)}\leq xy+xz+yz.$$
Thus, it's enough to prove that
$$16xyz(x+y+z)(xy+xz+yz)\leq\left(\sum_{cyc}(x^2y+x^2z+2xyz)\right)^2.$$
Now, let $\sum\limits_{cyc}(x^2y+x^2z)=6uxyz.$
Thus, by AM-GM $u\geq1$ and we need to prove that
$$16(6u+3)\leq(6u+6)^2$$ or $$(u-1)(3u+1)\geq0,$$ which is obvious.
