Đáp án:
$A_{min}=2$ khi $x=\dfrac{1}{2}$
Giải thích các bước giải:
Chờ bạn kia làm mà lâu quá :(
Áp dụng Bunhiacopxki:
$\left(x^2+\dfrac{1}{x} \right)\left(\dfrac{1}{4}+2 \right) \geq \left(\dfrac{x}{2}+\dfrac{\sqrt{2}}{\sqrt{x}} \right)^2$
$⇒x^2+\dfrac{1}{x} \geq \dfrac{4}{9} \left(\dfrac{x}{2}+\dfrac{\sqrt{2}}{\sqrt{x}} \right)^2$
$⇒\sqrt{x^2+\dfrac{1}{x}} \geq \dfrac{2}{3} \left(\dfrac{x}{2}+\dfrac{\sqrt{2}}{\sqrt{x}} \right)$
$⇒A \geq x+\dfrac{2}{3} \left(\dfrac{x}{2}+\dfrac{\sqrt{2}}{\sqrt{x}} \right)=\dfrac{1}{3} \left(4x+\dfrac{2\sqrt{2}}{\sqrt{x}} \right)$
$⇒A \geq \dfrac{1}{3} \left(4x+\dfrac{\sqrt{2}}{\sqrt{x}}+\dfrac{\sqrt{2}}{\sqrt{x}} \right)$
$⇒A \geq \dfrac{1}{3}.3\sqrt[3]{4x·\dfrac{\sqrt{2}}{\sqrt{x}}·\dfrac{\sqrt{2}}{\sqrt{x}}}=2$
$A_{min}=2$ khi $x=\dfrac{1}{2}$
Cách 2:
$x >0 \Rightarrow A>0$
$A-x=\sqrt{x^2+\dfrac{1}{x}}$
$⇒A^2-2Ax+x^2=x^2+\dfrac{1}{x}$
$⇒A^2=2Ax+\dfrac{1}{x} \geq 2\sqrt{2Ax·\dfrac{1}{x}}$
$⇒A^2 \geq 2\sqrt{2A}$
$⇒A^4 \geq 8A$
$⇒A^3 \geq 8$
$⇒A \geq 2$
$⇒A_{min}=2$ khi $x=\dfrac{1}{2}$
