证明1+xln(x+根号(x^2+1)>=根号(x^2+1)

f(x)=1+xln[x+√(x^2+1)]- √(x^2+1)
f'(x)=ln[x+√(x^2+1)]+x/√(x^2+1) -x/√(x^2+1)=ln[x+√(x^2+1)]
f'(-x)=ln[-x+√(x^2+1)]=-ln[x+√(x^2+1)]
x>0 时f'(x)>ln(1)=0
x