Prove that \(x > \ln x\) for all \(x > 0\).
<1>2. The function \(f\) has a global minimum at \(x=1\).
<1>3. Therefore \(f(x)\ge f(1)=1>0\) for every \(x>0\).
<1>4. Hence \(x>\ln x\) for every \(x>0\).
Solution
1. Define \(f:(0,\infty)\to\mathbb R\) by\[ f(x)=x-\ln x. \]
We have \[ f'(x)=1-\frac1x=\frac{x-1}{x}. \] Thus \(f'(x)<0\) for \(0<x<1\), \(f'(1)=0\), and \(f'(x)>0\) for \(x>1\). Hence \(f\) decreases on \((0,1]\) and increases on \([1,\infty)\), so its global minimum occurs at \(1\).
This follows from 2 and \(\ln 1=0\).
The inequality \(f(x)>0\) is exactly \(x-\ln x>0\).