Cho x>1. C/m: x+$\frac{1}{x-1}$ $\geq$ 3

$x>1\to x-1>0\\x+\dfrac{1}{x-1}\\=(x-1)+\dfrac{1}{x-1}+1$

Áp dụng BĐT Cô-si cho 2 số dương $x-1,\dfrac{1}{x-1}$ ta được:

$(x-1)+\dfrac{1}{x-1}\ge 2\sqrt{(x-1).\dfrac{1}{x-1}}=2\\\to (x-1)+\dfrac{1}{x-1}+1\ge 3\\\to x+\dfrac{1}{x-1}\ge 3$

Dấu "$=$" xảy ra khi: $x-1=\dfrac{1}{x-1}↔(x-1)^2=1↔x=2\text{(Tm)}$