hello, but $\frac{R}{R}=1$? Sonnhard.

Konigsberg wrote: Let $ABC$ be an obtuse triangle with $\angle A>90^{\circ}$, and let $r$ and $R$ denote its inradius and circumradius. Prove that $\frac{R}{R} \le \frac{asinA}{a+b+c}$ I think it would be $\frac rR$; I solved some similar inequaity before but cannot remember where;

oops, edited. :-(

Konigsberg wrote: Let $ABC$ be an obtuse triangle with $\angle A>90^{\circ}$, and let $r$ and $R$ denote its inradius and circumradius. Prove that $\frac{4}{R} \le \frac{asinA}{a+b+c}$ Are you sure $\frac 4R$ ??

hello, do you meant $\frac{r}{R}$ or $\frac{R}{r}$? Sonnhard.

You pasted it wrong. tensor wrote the correct version. So much negligence. Anyways the solution can be seen in hyperbolictangent's blog, Problem 1.18.

tensor wrote: $ABC$ is obtuse with $A\ge \frac {\pi}{2}$ consequently $B+C\le \frac{\pi}{2}$ So, $B$ and $C$ both are acute.....Now, WLOG Take $B\geq C$ so, $B+C\geq2C$ $\sin (B+C)\geq\sin 2C\geq {2\sin C\cos C}\geq {2\sin B\sin C}$ (since $\frac{\pi}{2}-C\geq B\implies \cos C\geq\sin B)$ $\begin{aligned}\sin (B+C)\geq {2\sin B\sin C\implies 1\geq\frac {2\sin B\sin C}{\sin (B+C)} \implies R\geq\frac {2R\sin B.2R\sin C}{2R\sin (B+C)}=\frac {2R\sin B.2R\sin C}{2R\sin A}\end{aligned}}$ since $\sin(B+C)=\sin A$ $\begin{align*}R\geq\frac {bc}{a}\implies bc\leq {aR}\implies 1\leq \frac {aR}{bc}\implies \Delta\leq \frac{aR\Delta}{bc}\implies \Delta\leq\frac {aR\(\frac {bc\sin A}{2}}{bc}$ Rearranges gives cancelling out the term $bc$ $\begin{align*}2\Delta\leq {aR\sin A}\implies 2rs\leq {aR\sin A}\implies r(a+b+c)\leq {aR\sin A}\implies \frac{r}{R}\leq\frac {a\sin A}{a+b+c}$ Dooone!!!!! Hey tensor remember this PM? I corrected some arguments and fixed up some LaTeX but the solution is tensor's. And I think its correct.

Thanks buddy.... but i think there is still a typo $\frac {\pi}{2}-B\geq C\implies \cos C\geq \sin B$ (as B,C acute and traking cosine the sign gets reversed) which is explicitly used in the previous line... anyways thanks again... here is a another link similar to this one... infact more than similar identical , after some manipulation http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&t=571618 And yeah infact the inequality is strict as mentioned in the blog or the link but in the post we got equality because we assumed $A\geq 90^{o}$ While in question it is $A>90^{o}$

No it is not a typo I am taking $\sin$ on both sides hence left side becomes $\cos C\ge \sin B$ . So it is ok, I think.

That's obviously right; But where did you use it?? .... $2\sin C\cos C\geq 2\sin B\sin C\iff \cos C\geq \sin B$... SO you need this staff ... right?

-_- Darn I need to learn how to reduce typos and also how to see my typos. Thanks bro I am editing it.

Okay, finally; this is the corrected version

The extended law of sines allows us to write the right hand side as $$\frac{a\sin A}{a + b + c} = \frac{1}{R} \cdot \frac{a^2}{2(a+b+c)},$$while the area formulas of $[ABC] = \tfrac{1}{2}bc\sin A = sr$ allow us to write the left hand side as $$\frac{r}{R} = \frac{1}{R}\cdot \frac{bc\sin A}{a+b+c}.$$Cancelling out the equal stuff, our inequality is thus equivalent to showing that $\tfrac{a^2}{2}\ge bc\sin A$, and this is not difficult: we have $$\frac{a^2}{2} \ge \frac{b^2+c^2}{2} \ge bc \ge bc \sin A,$$where the first inequality follows by $\angle A > 90^\circ$, the second by AM-GM, and the third by the fact that $\sin A \le 1$.