Use the arithmetic mean-geometric mean inequality to establish the following results:
(a) If $nt > -1 $, then $(1-t)^{n} \geq 1 - nt$
(b) If $-x < n < m$, then $(1+x/n)^{n} \leq (1 + x/m)^{m}$
MY ATTEMPT
(a) I had no success at trying to prove it using the AM-GM inequality. Thus I've tried to apply induction.
The base case is true, since for $n = 1$ we get that $(1-t)^{1} = 1 - t = 1 - 1\cdot t$.
Now let us assume that it holds for some natural number $n\geq 1$ and we shall prove it holds for $n + 1$ too.
Indeed, one has that \begin{align*} (1-t)^{n+1} = (1-t)^{n}(1-t) \geq (1-nt)(1-t) = 1 - (n+1)t + nt^{2} \geq 1 - (n+1)t \end{align*} and we are done.
(b) Similarly, we shall prove it applying induction on $m$. For $m = 2$ it is true, because \begin{align*} n < 2 \Rightarrow n = 1 \Rightarrow \left(1 + \frac{x}{2}\right)^{2} = 1 + x + \frac{x^{2}}{4} \geq 1 + x \end{align*}
But then I get stuck.
Could someone please finish the induction proof and provide a solution based on AM-GM inequality?
Any contribution is appreciated.