How to see that $e^{x(1+x/3)} \le (1+x)^{(1+x)}$ for very small x>0? I am reading a research paper and one of the calculation states that 

$e^{x(1+x/3)} \le (1+x)^{(1+x)}$ for very small x>0.

Is this true? How to prove it? Thank you very much.

Consider the equivalent $$f(x) = \log(1+x) - \frac{x}{1+x}\left(1+\frac{x}k\right) \ge 0$$
Now $f'(x) = \dfrac{x(k-2-x)}{(1+x)^2}$, so for any $k> 2$, the numerator remains positive for $0< x< k-2$ and $f(0)=0$, so the inequality holds for some nbd to the right of $0$. 
