令f(x)=ln(x+1)-2x/(x+2),x≥0则 f '(x)=1/(x+1) -4/(x+2)²=x²/[(x+1)(x+2)²]>0从而 f(x)在[0,+∞)是增函数所以 当x>0时,f(x)>f(0)=0即 ln(x+1)-2x/(x+2)>0ln(x+1)>2x/(x+2)