Ta có:
A=$\frac{x+3}{\sqrt[]{x}+1}$ =$\sqrt[]{x}$ -1 +$\frac{4}{\sqrt[]{x}+1}$ =$\sqrt[]{x}$ +1 +$\frac{4}{\sqrt[]{x}+1}$ - 2
Do x $\geq$ 0 => $\sqrt[]{x}$ +1 $\geq$ 1 > 0 và $\frac{4}{\sqrt[]{x}+1}$ > 0
Áp dụng BĐT Cô-si $\sqrt[]{x}$ +1 +$\frac{4}{\sqrt[]{x}+1}$ $\geq$ 2$\sqrt[]({\sqrt[]{x} +1 )}$ .$\frac{4}{\sqrt[]{x}+1}$ =2$\sqrt[]{4}$ =4
=>$\sqrt[]{x}$ +1 +$\frac{4}{\sqrt[]{x}+1}$ -2 $\geq$ 4 - 2=2
hay A $\geq$ 0
Dấu "=" xảy ra <=> $\sqrt[]{x}$ +1 = $\frac{4}{\sqrt[]{x}+1}$ <=> x=1
Vậy minA=2 khi x=1
