How to prove that $ 1- \frac{x^2}{n} \leq (1+\frac{x}{n})^n\cdotp(1-\frac{x}{n})^n$ How would I prove this inequality (assuming its true, its from a textbook)
$$1 - \frac{x^2}{n} \leq (1+\frac{x}{n})^n\cdotp(1+\frac{-x}{n})^n$$ 
if $n > |x|$, $x\in R$ and $n\in N$
I first rewrote the inequality to
$$1 - \frac{x^2}{n} \leq (1-\frac{x^2}{n^2})^n$$I then tried to manipulate the inequalities by saying the right hand side was greater than a smaller expression however I was unable to prove the above. I also tried induction where the base case works however I was unable to show that a case being true implies the next also being true.
Any help would be appreciated
 A: $$1-\frac{x^2}{n}\overset{?}{\leq}\left(1-\frac{x^2}{n^2}\right)^n$$
is a good starting point. You can assume $|x|<\sqrt{n}$, since otherwise the inequality is trivial, with the LHS being non positive and the RHS being positive.
Consider the logarithm of both sides. Then:
$$\log\left(1-\frac{x^2}{n}\right)\leq n\log\left(1-\frac{x^2}{n^2}\right)$$
is a consequence of the inequality:
$$\forall z\in[0,1),\qquad \log(1-z)\leq n\log\left(1-\frac{z}{n}\right)$$
that follows from the fact that:
$$\int_{0}^{z}\frac{dx}{1-x}\geq\int_{0}^{z}\frac{dx}{1-\frac{x}{n}}$$
since $(1-x)\leq 1-\frac{x}{n}$.
A: Deriving both sides on $x$, $$-\frac{2x}{n}\le-n\frac{2x}{n^2}(1-x^2)^{n-1},$$ or
$$-1\le-(1-\frac{x^2}{n^2})^{n-1}.$$
The latter relation is obviously true for $|x|<n$, so that the LHS of the initial relation decreases faster than the RHS, while they are equal for $x=0$.
(If you prefer, $l'(x)\le r'(x)\implies l'(x)-r'(x)\le0\implies l(x)-r(x)$ is decreasing $\implies l(x)-r(x)\le l(0)-r(0)=0\implies l(x)\le r(x)$.)
