\[ \prod \frac{a}{b+c}= \prod \frac{\sin{A}}{\sin{B}+\sin{C}} = \frac{\sin{\frac{A}{2}}\cos{\frac{A}{2}}}{\frac{\sin{B}+\sin{C}}{2}} \] $\sin{x}$ is concave on $[0, \pi]$, thus: \[ \frac{\sin{B}+\sin{C}}{2} \leq \sin{\frac{B+C}{2}}=\cos{\frac{A}{2}} \] Pluge it in we get the desired ineqaulity. Equality holds if and only if $\angle{A}=\angle{B}=\angle{C}$.

Lemma. $\triangle ABC\Longrightarrow \boxed {\ \sin \frac A2\le \frac{a}{b+c}\ }\ \ (*)\ .$ The first method. Define: the second intersection $A'$ between the circumcircle $C(O,R)$ and the bisector $[AI$ of the angle $\widehat {BAC}$; the intersection $D\in BC\cap AI$ $(l_a=AD)$. From the well-known relations $l_a=\frac{2bc}{b+c}\cdot \cos \frac A2$ and $AA'\cdot AD=bc$ $(ABA'\sim ADC)$ results: $AA'\le 2R$ $\Longrightarrow$ $bc\le 2Rl_a$ $\Longleftrightarrow$ $bc\le 2R\cdot \frac{2bc}{b+c}\cdot\cos\frac A2$ $\Longleftrightarrow$ $b+c\le 4R\cos \frac A2$ $\Longleftrightarrow$ $\frac{b+c}{a}\le 2\cdot \frac{\cos \frac A2 }{\sin A}$ $\Longleftrightarrow$ $\frac{b+c}{a}\le \frac{1}{\sin \frac A2}$ $\Longleftrightarrow$ $\sin \frac A2\le \frac{a}{b+c}\ .$ Therefore, the inequality $(*)$ is equivalently with $AA'\le 2R\ !$ The second method. Using the identity $bc=p(p-a)+(p-b)(p-c)$ results: $(*)\Longleftrightarrow \sqrt{\frac{(p-b)(p-c)}{bc}}\le \frac{a}{b+c}$ $\Longleftrightarrow$ $(p-b)(p-c)(b+c)^2\le a^2bc$ $\Longleftrightarrow$ $(p-b)(p-c)(b+c)^2\le a^2 [p(p-a)+(p-b)(p-c)]$ $\Longleftrightarrow$ $(p-b)(p-c)\left[(b+c)^2-a^2\right]\le a^2p(p-a)$ $\Longleftrightarrow$ $4p(p-a)(p-b)(p-c)\le a^2p(p-a)\Longleftrightarrow 4(p-b)(p-c)\le a^2\Longleftrightarrow$ $a^2-(b-c)^2\le a^2\Longleftrightarrow (b-c)^2\ge 0$, what is truly. Remark. Using this lemma, the proposed inequality is a immediate consequence. The equality holds if and only if $a=b=c$. Moreover prove easily that $\boxed {\frac 12\left(\frac rR\right)^2\le \boxed {\prod \sin \frac A2\le \prod \frac{a}{b+c}}\le \frac 18}\ .$

Well, I want to point some remarks here. This inequality is equivalent with \[ \frac{R}{r}\geq\frac{(a+b)(b+c)(c+a)}{4abc}. \] If I'm not mistaken this was proposed by Dorin Andrica in 1984 in Romanian Olympiad. Anyway, related to this one here is a stronger one, but all acute-angled triangle: \[ \frac{R}{r}\geq\sqrt{\frac{(a^{2}+b^{2})(b^{2}+c^{2})(c^{2}+a^{2})}{2a^{2}b^{2}c^{2}}}. \] This inequality has been proposed for the first time by Tudorel Lupu and Dumitru Anton in local olympiad in Constantza in 1988. In 1997, the same problem appears in American Mathematical Monthly due to Mihaly Bencze and Florin Popovici.

Thanks Cezar for Remarks.

Another solution:

Attachments:
Iran_1997_1.pdf (38kb)

Cezar Lupu wrote: Well, I want to point some remarks here. This inequality is equivalent with \[ \frac{R}{r}\geq\frac{(a+b)(b+c)(c+a)}{4abc}. \] If I'm not mistaken this was proposed by Dorin Andrica in 1984 in Romanian Olympiad. Anyway, related to this one here is a stronger one, but all acute-angled triangle: \[ \frac{R}{r}\geq\sqrt{\frac{(a^{2}+b^{2})(b^{2}+c^{2})(c^{2}+a^{2})}{2a^{2}b^{2}c^{2}}}. \] This inequality has been proposed for the first time by Tudorel Lupu and Dumitru Anton in local olympiad in Constantza in 1988. In 1997, the same problem appears in American Mathematical Monthly due to Mihaly Bencze and Florin Popovici. \[\frac{R}{r} \ge \frac{1}{2}\left(\frac{64a^2b^2c^2}{(4a^2-(b-c)^2)(4b^2-(c-a)^2)(4c^2-(a-b)^2)}\right)^2.\] (India tst 2006 p7) http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=485987

Amir.S wrote: Show that for any arbitrary triangle $ABC$, we have \[\sin\left(\frac{A}{2}\right) \cdot \sin\left(\frac{B}{2}\right) \cdot \sin\left(\frac{C}{2}\right) \leq \frac{abc}{(a+b)(b+c)(c+a)}.\] Show that for any arbitrary triangle $ABC$, we have \[\cos^2A+\cos^2B+\cos^2C\geq \frac{1}{2}(\frac{a^2}{b^2+c^2}+\frac{b^2}{c^2+a^2}.+\frac{c^2}{a^2+b^2}).\]

Sqing, a nice problem! Here is the solution:

Attachments:
Problem04072012.pdf (17kb)

vslmat wrote: Sqing, a nice problem! Here is the solution: Very Good. But ,only for acute triangle was founded.

Oh, Sqing, sorry for being so careless! The second part of the solution is here

Attachments:
Problem04072012_b.pdf (18kb)

$a^2=(c\cdot\cos B+b\cdot\cos C)^2\stackrel{\mathrm{(C.B.S)}}{\ \le\ }\left(c^2+b^2\right)\left(\cos^2B+\cos^2C\right)\implies$ $\sum\left(\cos^2B+\cos^2C\right)\ge \sum \frac {a^2}{b^2+c^2}\implies$ $\sum\cos^2A\ge\frac 12\cdot \sum \frac {a^2}{b^2+c^2}$ .

vslmat wrote: Oh, Sqing, sorry for being so careless! The second part of the solution is here Dear vslmat . Right triangle?

$\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}=\frac{\cosh+\cosh+\cosh-1}{4}$ Thus it suffices to prove that $\sum{\frac{a^2+b^2-c^2}{8ab}}-\frac{1}{4} \le \frac{abc}{(a+b)(b+c)(c+a)}$ $\Leftrightarrow \sum_{%Error. "cyclic" is a bad command. }{a^2b}+\sum_{%Error. "cyclic" is a bad command. }{ab^2}-\sum{a^3}-2abc \le \frac{8a^2b^2c^2}{(a+b)(b+c)(c+a)}$ Expanding both the sides and using Muirhead's theorem does the job!!

sqing wrote: Show that for any arbitrary triangle $ABC$, we have \[\cos^2A+\cos^2B+\cos^2C\geq \frac{1}{2}(\frac{a^2}{b^2+c^2}+\frac{b^2}{c^2+a^2}.+\frac{c^2}{a^2+b^2}).\] International Zhautykov Olympiad 2018/1

sqing wrote: Show that for any arbitrary triangle $ABC$, we have \[\cos^2A+\cos^2B+\cos^2C\geq \frac{1}{2}(\frac{a^2}{b^2+c^2}+\frac{b^2}{c^2+a^2}.+\frac{c^2}{a^2+b^2}).\] Proof of zhumazhenis: $$\cos^2A+\cos^2B=\frac{\left(c^2+b^2-a^2\right)^2}{4b^2c^2}+\frac{\left(c^2+a^2-b^2\right)^2}{4a^2c^2}\geq\frac{4c^4}{4c^2(a^2+b^2)}=\frac{c^2}{a^2+b^2}$$Very nice.

sqing wrote: Show that for any arbitrary triangle $ABC$, we have \[\cos^2A+\cos^2B+\cos^2C\geq \frac{1}{2}(\frac{a^2}{b^2+c^2}+\frac{b^2}{c^2+a^2}.+\frac{c^2}{a^2+b^2}).\] it was here http://artofproblemsolving.com/community/c6h423450p2394999

mudok wrote: sqing wrote: Show that for any arbitrary triangle $ABC$, we have \[\cos^2A+\cos^2B+\cos^2C\geq \frac{1}{2}(\frac{a^2}{b^2+c^2}+\frac{b^2}{c^2+a^2}.+\frac{c^2}{a^2+b^2}).\] it was here http://artofproblemsolving.com/community/c6h423450p2394999 Yeah.

\(\prod \frac{a}{b+c}= \prod \frac{\sin{A}}{\sin{B}+\sin{C}}\) Could you explain how you can conclude this?

Evenprime123 wrote: \(\prod \frac{a}{b+c}= \prod \frac{\sin{A}}{\sin{B}+\sin{C}}\) Could you explain how you can conclude this? It is just the sine rule