Prove $\ln(1+x)\geq x-\frac{x^2}{2}$

Let $f(x)=\ln(1+x)-x+\frac{x^2}{2}$.

Thus, $$f'(x)=\frac{1}{x+1}-1+x=\frac{x^2}{1+x}\geq0.$$

Thus, $f(x)\geq f(0)=0$ and we are done!

Observe that $$\ln(1+x) = \int_0^{x} \frac{1}{1+t}dt.$$ But, for $t \ge 0$, $$\frac{1}{1+t} \ge 1-t.$$ Therefore, $$\ln(1+x) \ge \int_{0}^x(1-t)dt=x-\frac{x^2}{2}.$$

You want to show $$1+x\geq e^{x-x^2/2}$$

or $e^{x^2/2}(1+x)\geq e^{x}.$

The right side is $\sum_{k=0}^{\infty} \frac{x^k}{k!}$ and the left side is:

$$(1+x)\sum_{j=0}^{\infty}\frac{x^{2j}}{2^jj!}=\sum_{k=0}^{\infty}\frac{x^{k}}{2^{\lfloor k/2\rfloor}\lfloor k/2\rfloor!}$$

But $2^{\lfloor k/2\rfloor}\lfloor k/2\rfloor!\leq k!$.