Question

Answer 1

\(\dfrac{1}{P}-\dfrac{\sqrt{a}+1}{8}\ge1\left(a>0\right)\\ \Leftrightarrow\dfrac{2\sqrt{a}}{\sqrt{a}+1}-\dfrac{\sqrt{a}+1}{8}-1\ge0\\ \Leftrightarrow\dfrac{16\sqrt{a}-a-2\sqrt{a}-1-8\sqrt{a}-8}{8\left(\sqrt{a}+1\right)}\ge0\\ \Leftrightarrow-a+6\sqrt{a}-9\ge0\\ \Leftrightarrow-\left(\sqrt{a}-3\right)^2\ge0\)

Mà \(-\left(\sqrt{a}-3\right)^2\le0,\forall a\)

Do đó \(\sqrt{a}-3=0\Leftrightarrow a=9\left(tm\right)\)