Prove simple inequality probably with exponentials

Question

I can't prove

$$\frac{1+nx^2}{(1+x^2)^n} \leq 1$$

for $x,n>0 $

I know this is an easy question, but no matter how many exponential inequalities I try to use I can't crack it. Is there some sort of trick?

Answer 1

The trick is not to use exponential inequalities. Instead, use the binomial expansion : $$(a+b)^n = \sum_{i=0}^{n} \binom{i}{n} a^ib^{n-i}$$ With $a=1$ and $b=x^2$, you get $$(1+x^2)^n = 1 + n x^2 + \underbrace{\sum_{i=2}^{n} \binom{i}{n} (x^2)^{n-i}}_{\geqslant 0}$$

Hence, $(1+x^2)^n\geqslant 1 + n x^2$.

Answer 2

If $n$ is supposed to be a postive integer then $(1+x^{2})^{n}=1+nx^{2}+\cdots +x^{2n}$ by Binomial Theorem. Hence $(1+x^{2})^{n} \geq 1+nx^{2}$ which is equivalent to your inequality.

Answer 3

Hint: Use Bernoulli's inequality.