This is really interesting Namely, before 2 weeks we had Balkan M.O. preparation test and question equivalent to your inequality (related to medians and sidelengths) was proposed and as I remember we could not solve it by using some standard methods but in Sharygin's book there is non/standard solution to this inequality. I will try to find a book and scan that part.

Thanks delegat for the information. I 'll try it before looking at such an elegant solution.

delegat wrote: If $ a$, $ b$ and $ c$ are positive reals prove inequality: $ (1 + \frac {4a}{b + c})(1 + \frac {4b}{a + c})(1 + \frac {4c}{a + b}) > 25$ It's a privet case of my inequality. See here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=59512

mathangel Here I attached scanned pages with solution to geometric inequality written above.

Thanks for sharing. While looking for the solution, I transformed the problem to another one: Given triangle $ ABC$ whose sides are $ a,b,c$ and whose circumradius is $ R$. Denote by $ R_m$ the radius of the circumcircle of the triangle formed by the 3 medians of $ \triangle ABC$. Then we have the inequality: $ R_m \geq \frac{5}{6} R$. It seems that this is the original problem in the book, isn't it ? Anyway, the solution is elegant.

You are absolutely right mathangel The same question as you wrote is asked in that book.

this the problem of the gyumri olymiad in 2008, when I was in 7class, in our class no one solved, and after seeing the solution we didn't understant anything, but now it is very easy after opening the brackets we get $a^3+b^3+c^3+12abc>a^2b+b^2a+c^2a+a^2c+c^2b+b^2c$ which is true by chur's ineq, which is $a^3+b^3+c^3+3abc\ge a^2b+b^2a+c^2a+a^2c+c^2b+b^2c=>$ PROOVED

arshakus wrote: this the problem of the gyumri olymiad in 2008, when I was in 7class, in our class no one solved, and after seeing the solution we didn't understant anything, but now it is very easy after opening the brackets we get $a^3+b^3+c^3+12abc>a^2b+b^2a+c^2a+a^2c+c^2b+b^2c$ which is true by chur's ineq, which is $a^3+b^3+c^3+3abc\ge a^2b+b^2a+c^2a+a^2c+c^2b+b^2c=>$ PROOVED Bosnia Herzegovina Regional Olympiad 2008: $ \left(1+\frac{4a}{b+c}\right)\left(1+\frac{4b}{a+c}\right)\left(1+\frac{4c}{a+b}\right) > 25$ $\Leftrightarrow a^3+b^3+c^3+7abc>a^2b+b^2a+c^2a+a^2c+c^2b+b^2c$

Book: Inegalitati; Authors:L.Panaitopol,V. Bandila,M.Lascu http://www.artofproblemsolving.com/Forum/posting.php?mode=quote&f=151&t=569686&p=3344462: Let $a,b,c$ be positive real numbers . Prove that\[(1+\frac{6a}{b+c})(1+\frac{6b}{c+a})(1+\frac{6c}{a+b})> 43.\] http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=496088&p=2785707#p2785707 http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&p=2685748#p2685748 http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&p=3344407#p3344407

Regional Olympiad - Federation Of Bosnia And Herzegovina 2008 delegat wrote: If $ a$, $ b$ and $ c$ are positive reals prove inequality: \[ \left(1+\frac{4a}{b+c}\right)\left(1+\frac{4b}{a+c}\right)\left(1+\frac{4c}{a+b}\right) > 25.\] If $ a$, $ b$ and $ c$ are positive reals prove inequality: $$ ( \frac{4a}{b+c}+1)( \frac{4b}{c+a}+1)( \frac{4c}{a+b}+1)\geq 25+\frac{16abc}{(a+b)(b+c)(c+a)}. $$

sqing wrote: If $ a$, $ b$ and $ c$ are positive reals prove inequality: $$ ( \frac{4a}{b+c}+1)( \frac{4b}{c+a}+1)( \frac{4c}{a+b}+1)\geq 25+\frac{16abc}{(a+b)(b+c)(c+a)}. $$ 7436.

Bravo!!!!!