Using Bernoulli's Inequality show that $$\left(1+\frac{1}{k+1}\right)\left(1-\frac{1}{(k+1)^2}\right)^k\geq 1$$for all $k\in\mathbb{Z^+}$.
My initial thought was to start be noting that \begin{align*}1-\frac{1}{(k+1)^2}&\geq 1-\frac14 \text{ for all natural $k$}\\\implies\left(1-\frac{1}{(k+1)^2}\right)^k&\geq\left(1-\frac14\right)^k\geq1-\frac k4,\end{align*}by Bernoulli's Inequality. Multiplying both sides by $(1+1/(k+1))$ gives the required expression on the left hand side, but the expression on the right hand side is not always greater than or equal to one. So my approach is clearly flawed.
However, if I use Bernoulli's Inequality as follows $$\left(1-\frac{1}{(k+1)^2}\right)^k\geq1-\frac{k}{(k+1)^2}$$ and multiply both sides by $(1+1/(k+1))$ I get an expression greater than or equal to one. However, this seems to me completely false; isn't Bernoulli's inequality $$(1+x)^k\geq1+xk$$ supposed to work for a variable real $x>-1$ and a fixed natural $n$?
Any help will be appreciated.
Best Answer
Yes, notice that $(k+1)^2 \geqslant 1$, and thus we are allowed to use the Bernoulli's inequality. Then $\left(1 -\frac{1}{(k+1)^2}\right)^k \geqslant 1 -\frac{k}{(k+1)^2}$. It is enough to show that $(1+\frac{1}{k+1})(1-\frac{k}{(k+1)^2})\geqslant 1$. After multiplication and canceling out the necessary terms, the last inequality is equivalent to $\frac{1}{(k+1)^3} > 0$, which is true.