1. x,y>0. Tìm$A_{min}=\frac{(x+y)^{2}}{(x^{2}+y^{2})}+\frac{(x+y)^{2}}{xy}$
2. a,b,c>0; a+b+c=1. Tìm $B_{max}=\frac{ab}{\sqrt{c+ab}}+\frac{bc}{\sqrt{a+bc}}+\frac{ac}{\sqrt{b+ac}}$
sai đề mà –  ๖ۣۜDevilღ 25-07-15 08:58 PM
đúng rồi ah. a Khờ giúp e bài 2 luôn dk ko? –  Đá Nhỏ 25-07-15 08:49 AM
đúng đề chưa vậy nè –  ๖ۣۜDevilღ 24-07-15 11:04 PM
$\frac{(x+y)^2}{x^2+y^2}+\frac{(x+y)^2}{xy}=\left ( x+y \right )^2\left ( \frac{1}{x^2+y^2}+\frac{1}{2xy} +\frac{1}{2xy}\right )\geq \left ( x +y\right )^2\left ( \frac{4}{(x+y)^2}+\frac{1}{2xy} \right )=4+\frac{(x+y)^2}{2xy}\geq 4+\frac{4xy}{2xy} =6$
$x=y$

Cần trả +1,000vỏ sò để xem nội dung lời giải này
@@ thế đăng bài đi em ...:P –  Thu Cúc 25-07-15 05:06 PM
ơ, Max=1/2 mà. mà đề là ab/căn(c ab) chứ có phải là căn (ab/(c ab))đâu ạ –  Trần Minh Hoàng 25-07-15 04:40 PM

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