How to prove this lower bound on $\log(1+x)$ for $x \geq 0$? How about $x > -1$?

Tbe problem looks complex and encourages us to use approximations and taylor series but its infact very easy

Consider $$f(x)=\ln(1+x)-\frac{x(5x+6)}{2(x+3)(x+1)}$$ Indeed (after some hardwork)$$f'(x)=\frac{x^3}{{(x+3)}^2{(x+1)}^2}\ge 0$$

Thus $$f(x)\ge f(0)=0$$ Done