parmenides51 wrote: Let $ABC$ be a triangle whose side lengths are opposite the angle $A,B,C$ are $a,b,c$ respectively. Prove that $$\frac{ab\sin{2C}+bc\sin{ 2A}+ca\sin{2B}}{ab+bc+ca}\leq\frac{\sqrt{3}}{2}$$. (nooonuii) stronger is $$\frac{ab\sin{2C}+bc\sin{ 2A}+ca\sin{2B}}{ab+bc+ca}\leq \frac{1}{3}\sum{\cos{\frac{A}{2}}}$$

Using $\sin(2A)=2\sin A\cos A$ and $a/\sin A=2R$, it boils down provving \[ \frac{abc}{R}\frac{\sum \cos A}{\sum ab}\le \frac{\sqrt{3}}{2} \iff \frac{\sum \cos A}{R(1/a+1/b+1/c)}\le \frac{\sqrt{3}}{2}. \]Notice that $\sum R/a = \frac12\sum(\sin A)^{-1}$. As a result, it boils down proving \[ \frac{\sum \cos A}{\sum (\sin A)^{-1}} \le \frac{\sqrt{3}}{4}. \]This is not so hard. Recalling the identity $\sum \cos A = 1+4\sin(A/2)\sin(B/2)\sin(C/2)$, the fact $\sin A+\sin B+\sin C\le 3\sqrt{3}/2$ (thanks to Jensen), as well as AM-GM; one can easily clinch above.

xzlbq wrote: parmenides51 wrote:

Let $ABC$ be a triangle whose side lengths are opposite the angle $A,B,C$ are $a,b,c$ respectively. Prove that (nooonuii)

. stronger is $$\frac{ab\sin{2C}+bc\sin{ 2A}+ca\sin{2B}}{ab+bc+ca}\leq \frac{1}{3}\sum{\cos{\frac{A}{2}}}$$ I'd like to say that this one is, in effect, not strong. Well, the stronger inequalities: \[\frac{r}{\sqrt3R}\sum_{\rm cyc}{\frac{g_a}{h_a}}\leqslant\frac{bc\sin{\mspace{-1.5mu}2A}+ca\sin{\mspace{-1.5mu}2B}+ab\sin{\mspace{-1.5mu}2C}}{bc+ca+ab}\leqslant\frac92\prod_{\rm cyc}{\tan{\mspace{-1.5mu}\frac{A}2}}\textnormal.\]Because $$\frac{4{\left(\sum{a}\right)}^3{\left(\sum{bc}\right)}^2\prod{bc}}{\prod{\left(b+c-a\right)}}{\left(\frac9{2}\prod{\tan{\mspace{-1.5mu}\frac{A}2}}-\frac{\sum{bc\sin{\mspace{-1.5mu}2A}}}{\sum{bc}}\right)}=3\sum{a{\left(a^2-ab-ac+2bc\right)}^2{\left(b-c\right)}^2}+12\prod{a}\sum{a^2{\left(a-b\right)}{\left(a-c\right)}}+\sum{bc}\sum{{\left(a-b\right)}{\left(a-c\right)}}\sum{a{\left(a-b\right)}{\left(a-c\right)}}\geqslant0\text.$$