Question: Is inequality $$\Big (1+\frac{1}{k(k+1)}\Big )^k \geq 1+\frac{1}{k+1}$$ is true for all $k \in \mathbb{N}$ ? I am proving this using binomial theorem on the right side but I don't know what to do in the middle since there are many terms and there is an exponent of $k$. Is there a better way to prove this one? Thanks. I think of using $AM \geq GM$ inequality but I don't know how to apply it.Thanks Again.

Answer: 
$$
\left(1+\frac{1}{k(k+1)}\right)^k=\sum\_{i=0}^k{k\choose i}\frac{1}{(k(k+1))^i}=1+\frac{1}{k+1}+\text{ something nonnegative} \ge 1+\frac{1}{k+1}.
$$
**Hint** Bernoulli inequality and see Calvin`s comment......

You need the stronger version of Bernoulli, namely the one which says "*with equality if and only if...*"