Đáp án:
Giải thích các bước giải:
Buniacopxki:
$\left ( 1.x+2\sqrt{2}.\dfrac{1}{\sqrt{x}} \right )^2\leq \left ( 1+8 \right )\left ( x^2+\dfrac{1}{x} \right )$
$⇒\sqrt{x^2+\dfrac{1}{x}} \geq \dfrac{1}{3}\left ( x+\dfrac{2\sqrt{2}}{\sqrt{x}} \right )$
$⇒P \geq \dfrac{4x}{3}+\dfrac{2\sqrt{2}}{3\sqrt{x}}$
$⇒P \geq \dfrac{2}{3}\left ( 2x+\dfrac{\sqrt{2}}{2\sqrt{x}}++\dfrac{\sqrt{2}}{2\sqrt{x}} \right )\geq \dfrac{2}{3}.3\sqrt[3]{\frac{2x.\sqrt{2}.\sqrt{2}}{2\sqrt{x}.2\sqrt{x}}}=2$
$P_{min}=2$ khi $x=\dfrac{1}{2}$
