bdt

 cho a,b,c >0
cmr $\frac{19b^{3}-a^{3}}{ab+5b^{2}}+\frac{19c^{3}-b^{3}}{cb+5c^{2}}+\frac{19a^{3}-c^{3}}{ac+5a^{2}}\leq 3(a+b+c)$
Ta CM $\frac{19b^3-a^3}{ab+5b^2}\le -a+4b \Leftrightarrow (-a+4b)(ab+5b^2)\ge 19b^3-a^3\Leftrightarrow (a-b)^2(a+b)\ge0$
 Tương tự cộng lại ta có đpcm
anh oi cho em hoi cach lam the nao de ra he so nhu tren –  kaito bing 24-03-17 06:56 AM
có nh cách,có thể đặt hệ số cái bt kia $\le \alpha a \beta b$ với $\alpha \beta =3$ hoặc cách thứ 2 là xuất phát từ $a^3 b^3\ge ab(a b)$ rồi biến đổi theo tử và mẫu –  ๖ۣۜPXM๖ۣۜMinh4212♓ 12-03-17 07:54 AM
có cơ sở nào để nghĩ đc cách này a?sao a lại đoán đc mà c/m –  ๖ۣۜTQT☾♋☽ 11-03-17 04:52 PM

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