Chứng minh:$a^2-\dfrac{3}{4a}-\dfrac{a}{1-a} \le \dfrac{-9}{4}$\Leftrightarrow $(2a-1)(a^2+3) \ge 0$ (luôn đúng) (Quy đồng)Mà $b \le 1-a$ nên $\dfrac{a}{b} \ge \dfrac{a}{1-a}$\Rightarrow đpcm.
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