Show that $\left(1+\frac xk\right)^k$ is monotonically increasing Note: I don't want to get the full solution, but only a hint.
I have to show that for $x \in [0, \infty)$, the sequence $\left(1+\frac xk\right)^k$ is monotonically increasing. We were already given the hint that we could use the inequality of arithmetic and geometric means, but I don't see how to apply it yet. 
I tried to do it by induction. While the case $k = 1$ works easily, I don't know how to start further from here. I already tried to use the definition of the binomial theorem, but it didn't lead me anywhere.
 A: Hint: 
Apply AM - GM inequality with $m > k$ to:
$$\left(1+\frac{x}{k}\right)^{k/m} =\left(\frac{k+x}{k}\right)^{k/m} = \left[\left(\frac{k+x}{k}\right)^k \right]^{1/m} \\= \left[\underbrace{\frac{k+x}{k} \ldots \frac{k+x}{k}}_k \underbrace{1 \ldots 1}_{m-k}\right]^{1/m} \\ \leqslant \frac{1}{m}\left(k \frac{k+x}{k } + (m-k)(1) \right)$$
A: The AM-GM tells us that for non negative $\;a_n\;$ :
$$\frac{a_1+\ldots+a_n}n\ge\sqrt[n]{a_1\cdot\ldots\cdot a_n}$$
Let us apply this with
$$a_1=1\,,\,\,a_2=a_3=\ldots=a_{n+1}=1+\frac xn\;,\;\;\text{and then get :}$$
$$\frac{a_1+\ldots+a_{n+1}}{n+1}\ge\sqrt[n+1]{a_1\cdot\ldots\cdot a_{n+1}}\iff$$
$$\frac{1+\left(1+\frac xn\right)+\ldots+\left(1+\frac xn\right)}{n+1}\ge\sqrt[n+1]{1\cdot\left(1+\frac xn\right)\cdot\ldots\cdot\left(1+\frac xn\right)}\stackrel{\text{raise to power}\;n+1 }\iff$$
$$\left(\frac{1+n\left(1+\frac xn\right)}{n+1}\right)^{n+1}\ge\left(1+\frac xn\right)^n\iff\left(1+\frac x{n+1}\right)^{n+1}\ge\left(1+\frac xn\right)^n$$
A: Expand it using the binomial theorem and try to look at the sum term by term. Create a sequence of each term (other than 1 obviously )and try to get that  it  increases for increasing $k $. 
The sum will be $1+^kC_1(\frac xk) + ^kC_2 (\frac xk)^2 + ... ^kC_k (\frac xk)^k $.This could be written as $1+\sum_{r=1}^k {^kC_r(\frac xk)^r}$ Now consider the sequence $^kC_r(\frac 1k)^r$ and try to prove that this sequence is increasing for increasing $k$ for any $r >=1$. This will prove that each term of the sum increases thus implying the sum increases .
One way to prove that the sequence $ S = ^kC_r(\frac 1k)^r$ increase is to calculate the ratio $S (k+1)/S (k) $ and prove that is greater than or equal to 1 for all $k $.
A: For $x=0$ the function is constant, so we can assume $x>0$.
Consider
$$
f(t)=\frac{\log(1+xt)}{t}
$$
for $t>0$. Then
$$
f'(t)=\frac{t\frac{x}{1+xt}-\log(1+xt)}{t^2}=
\frac{xt-(1+xt)\log(1+xt)}{t^2(1+xt)}
$$
Now our task is proving that, for $u>0$,
$$
g(u)=u-(1+u)\log(1+u)<0
$$
We indeed have $g(0)=0$ and
$$
g'(u)=1-\log(1+u)-1=-\log(1+u)
$$
Put together the pieces and conclude.
