View source for X/(1+x) less than log(1+x)

==Theorem==
The following formula holds for $x>-1$:
$$\dfrac{x}{1+x} < \log(1+x),$$
where $\log$ denotes the [[logarithm]].

==Proof==

==References==
* {{BookReference|Handbook of mathematical functions|1964|Milton Abramowitz|author2=Irene A. Stegun|prev=Euler-Mascheroni constant|next=log(1+x) less than x}}: $4.1.33$

[[Category:Theorem]]
[[Category:Unproven]]