Cho hai số thực dương thỏa: 2(a2+b2)+ab=(a+b)(ab+2)

Tìm GTNN của P=4(a3b3+b3a3)9(a2b2+b2a2)
Ta có: 2(a2+b2)+ab=(a+b)(ab+2)t=a2+b2ab=S2+SP12(S24P)
tS2+4S122212
t3=a3b3+b3a3+3(ab+ba);t2=a2b2+b2a2+2
f(t)=4t39t212t+12
f(t)=12t218t12t=2<2212 (loại);
f(t)f(2212)=...
Dấu bằng có khi S2=8;S22PP=2212
hiểu khúc đó r :) do nãy ko nghĩ ra đặt cái pt x^2 -S.x P=0 :) –  NguyễnTốngKhánhLinh 09-04-14 10:33 PM
khi đặt S,P thì điều kiện để có a,b thỏa là S2 ge4P mà. quên rồi hả –  ♂Vitamin_Tờ♫ 09-04-14 09:20 PM
Mon ơi, dòng đầu tiên, cái điều kiện là sao? –  NguyễnTốngKhánhLinh 09-04-14 09:06 PM

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