Đặt $t=\sqrt{x}+\dfrac{1}{\sqrt{x}}$
Theo BĐT AM-GM:
$t\ge 2\sqrt{ \sqrt{x}.\dfrac{1}{\sqrt{x}} }=2$
$\to t\ge 2$
$t^2=x+\dfrac{1}{x}+2$
$\to x+\dfrac{1}{x}=t^2-2$
$\to A=t^2-2-3t=t^2-3t-2$
• Chứng minh hàm $A(t)$ đồng biến khi $t\ge 2$:
$A(t)=t^2-3t-2$
$A(t+1)=(t+1)^2-3(t+1)-2=t^2-3t-2+2t-2$
$\to A(t+1)-A(t)=2t-2=2(t-1)>0\quad\forall t\ge 2$
$\to A(t+1)>A(t)\quad\forall t\ge 2$
$\to$ đpcm
Do đó $\min A=A(2)=-4$
Dấu $=$ xảy ra khi $\sqrt{x}+\dfrac{1}{\sqrt{x}}=2$
$\to x=1$
