Ta có:$A=\Sigma \frac{\frac{1}{a}+\frac{1}{b}}{\frac{2}{a}-\frac{1}{b}}$
Mà : $ \frac{1}{a} + \frac{1}{c} = \frac{2}{b}. Nên \frac{1}{a} + \frac{1}{b} = \frac{3}{b} - \frac{1}{c}$
Và: $ \frac{2}{a} - \frac{1}{b} = \frac{3}{b}-\frac{2}{c}$
Nên, $A=\Sigma \frac{\frac{3}{b}-\frac{1}{c}}{\frac{3}{b}-\frac{2}{c}}$
Đặt $\frac{3}{b}=x;\frac{1}{a}=y;\frac{1}{c}=z$
Có: $3y+3z=2x$
$A=\Sigma \frac{x-y}{x-2y}$
$2A=\Sigma \frac{2x-2y}{x-2y}$
$2A-2=\Sigma \frac{x}{x-2y}$
$2A-2= \frac{x}{x-2y} + \frac{x}{x-2z} \geq \frac{x.(1+1)^2}{2x-2y-2z}$
$= \frac{6y+6z}{y+z}=6$
Nên, $2A\geq 8 \Leftrightarrow A\geq 4$
Dấu = xảy ra $\Leftrightarrow y=z \Leftrightarrow a=c$
$\Leftrightarrow a = b = c$