$ \displaystyle \sum_{cyc}{\frac{(ab)^2}{c^3(a^2-ab+b^2)}} \ge \frac{3}{\sum_{cyc}{ab}} $
(proposed by a friend...)
un altra disuguaglianza (sarà piu facile della AM-QM?naaaaa)
un altra disuguaglianza (sarà piu facile della AM-QM?naaaaa)
Dati $ a,b,c>0 $ tali che $ \sum_{cyc}a=1 $ mostrare che:
$ \displaystyle \sum_{cyc}{\frac{(ab)^2}{c^3(a^2-ab+b^2)}} \ge \frac{3}{\sum_{cyc}{ab}} $
(proposed by a friend...)
$ \displaystyle \sum_{cyc}{\frac{(ab)^2}{c^3(a^2-ab+b^2)}} \ge \frac{3}{\sum_{cyc}{ab}} $
(proposed by a friend...)
The only goal of science is the honor of the human spirit.