Ta có 4ab-a-b=2 \Leftrightarrow a+b=4ab-2Từ BĐT cosi a+b \geq 2\sqrt{ab}\Leftrightarrow 4ab-2 \geq 2\sqrt{ab}\Leftrightarrow 2ab-\sqrt{ab}-1 \geq 0\Leftrightarrow \left ( \sqrt{ab}-1 \right )\times \left ( \sqrt{ab}+\frac{1}{2} \right ) \geq 0\Leftrightarrow \begin{cases}\sqrt{ab}\leq -\frac{1}{2}(loại vì \sqrt{ab} luôn dương)\\ \sqrt{ab}\geq 1 \end{cases}\Rightarrow a+b \geq 2\sqrt{ab} \geq 2\Rightarrow \frac{1}{a+b} \geq \frac{1}{2} \Rightarrow A\geq \frac{5}{2}dấu "=" xảy ra khi a=b=1

Ta có 4ab-a-b=2 $\Leftrightarrow$ a+b=4ab-2Từ BĐT cosi a+b $\geq$ 2$\sqrt{ab}$$\Leftrightarrow$ 4ab-2 $\geq$ 2$\sqrt{ab}$$\Leftrightarrow$ 2ab-$\sqrt{ab}$-1 $\geq$ 0$\Leftrightarrow$ ($\sqrt{ab}$ -1)$\times$($\sqrt{ab}$ +$\frac{1}{2}$) $\geq $ 0 $\Leftrightarrow$ \begin{cases}\sqrt{ab}\leq -\frac{1}{2}(loại vì \sqrt{ab} luôn dương)\\ \sqrt{ab}\geq 1 \end{cases}$\Rightarrow$ a+b $\geq$ 2$\sqrt{ab}$ $\geq$ 2$\Rightarrow$ $\frac{1}{a+b}$ $\geq$ $\frac{1}{2}$ $\Rightarrow$ A$\geq$ $\frac{5}{2}$dấu "=" xảy ra khi a=b=1