Bài tập đại 9^^

C

congchuaanhsang

Xét hiệu:
A=$\dfrac{2}{a^2+ab+b^2}$+$\dfrac{1}{3b^2}$-$\dfrac{9}{(a+2b)^2}$
=$\dfrac{6b^2(a^2+4ab+ab^2)+(a^2+ab+b^2)(a^2+4ab+4b^2)-27b^2(a^2+ab+b^2)}{3b^2(a^2+ab+b^2)(a+2b)^2}$
=$\dfrac{a^4+b^4+5ab^3+5a^3b-12a^2b^2}{3b^2(a^2+ab+b^2)(a+2b)^2}$
=$\dfrac{(a^4-2a^2b^2+b^4)+5ab(a^2-2ab+b^2)}{3b^2(a^2+ab+b^2)(a+2b)^2}$
=$\dfrac{(a^2-b^2)^2+5ab(a-b)^2}{3b^2(a^2+ab+b^2)(a^2+4ab+4b^2)}$
Vì a,b dương\RightarrowA\geq0
\Leftrightarrow$\dfrac{2}{a^2+ab+b^2}$+$\dfrac{1}{3b^2}$\geq$\dfrac{9}{(a+2b)^2}$
Dấu "=" xảy ra \Leftrightarrow a=b
 
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