For all nonnegative real numbers $a,b$ and $c.$ 
Prove that:
             $\color{blue}{\sum_{cyc}a^2\sum_{cyc}a(b+c)\sum_{cyc}\frac{1}{(b+c)^2}\geq (a+b+c)^4}$
Solution:
       $\sum_{cyc}\frac{1}{(a+b)^2}\geq \frac{3(a+b+c)^2}{8(ab+bc+ca)}(\frac{1}{ab+bc+ca}+\frac{1}{a^2+b^2+c^2})$

. . –  Confusion 10-08-16 08:55 PM
k rõ đề sao làm đey :v –  tran85295 10-08-16 08:31 PM
đề rõ thế còn rỳ, full j nữa tr @@ –  Confusion 10-08-16 08:02 PM
đây cx ko rõ :-/ –  Confusion 10-08-16 07:59 PM
chịu anh vs chả em! Vietnamese còn chưa học hết đã English! –  ★·.·´¯`·.·★Poseidon★·.·´¯`·.·★ 10-08-16 07:55 PM
đề nghị thím viết full đề :v –  ๖ۣۜPXM๖ۣۜMinh4212♓ 10-08-16 07:46 PM
ko rõ chỗ $\sum_{cyc}a(b \dotplus c$ là ntn nhỉ –  tran85295 10-08-16 07:42 PM
ơ, không đồng bậc à, căng à nha –  ๖ۣۜPXM๖ۣۜMinh4212♓ 10-08-16 07:39 PM

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