Show that $\sqrt[k]{a}\leq a_{n+1}\leq a_n$ for $n\in \mathbb{N}$ let $a\in \mathbb{R}:a>0$ and $k\in \mathbb{N}_{\geq 2}$. Furthermore $(a_n)_{n\geq 1} \subset \mathbb{R}$: $$a_1=1+a \text{ and } a_{n+1}=a_n\left(1+\frac{a-a_n^k} {ka_n^k}\right) \text{ for } n\in \mathbb{N}$$
Now I have to show that 

$$\sqrt[k]{a}\leq a_{n+1}\leq a_n$$ 

for $n\in \mathbb{N}$. I have been looking at this problem for quite some time now, and I don't understand what's going on here. How do I know what $a_n$ even is? And where do I even start with the proof? I have never encountered that kind of sequence.
 A: let $a\in \mathbb{R}:a>0$ and $k\in \mathbb{N}_{\geq 2}$. Furthermore $(a_n)_{n\geq 1} \subset \mathbb{R}$: $$a_1=1+a \text{ and } a_{n+1}=a_n\left(1+\frac{a-a_n^k} {ka_n^k}\right) \text{ for } n\in \mathbb{N}$$
We wish to show that $\forall n \in \mathbb N$:
$$\sqrt[k]{a}\leq a_{n+1}\leq a_n$$ 
We will induct on $n$,  let's start with $n=1$, we need to show that:
$$\sqrt[k]{a}\leq a_{2}\leq a_1$$
Observe indeed that this means: 
$$a_2=a_1\left(1+\frac{a-a_1^k} {ka_1^k}\right)=(1+a)\left(1+\frac{a-(1+a)^k} {k(1+a)^k}\right)=(1+a)(1 + \frac{a}{k(1+a)^k} - \frac{1}{k})$$
we know that $k\geq 2$ so $\frac{1}{k } \leq 2$ and since $a>0$ also $ \frac{1}{(1+a)^k} \leq \frac{1}{(1+a)^2}$
$$a_2 \leq (1+a)(1+ \frac{a}{2(1+a)^2} -\frac{1}{2})=(1+a) + \frac{a}{2(1+a)}-\frac{1}{2} $$ $$=(1+a) + \frac{1}{2(\frac{1}{a}+1)}-\frac{1}{2}<(1+a)+ \frac{1}{2}- \frac{1}{2}=1+a_1$$
so $a_2 \leq a_1$. Now we also need to prove that $\sqrt[k]{a} \leq a_2$ or $a \leq a_2^k$, we get:
$$a_2^k=(1+a)^k(1 + \frac{a}{k(1+a)^k} - \frac{1}{k})^k > (1+a)^k(1+ \frac{a}{k}-\frac{1}{k})^k = (1+a+\frac{a}{k} +\frac{a^2}{k}-\frac{a}{k}-\frac{1}{k})^k $$ $$a_2^k > (1+a+\frac{a^2}{k}-\frac{1}{k})^k $$
We now use that $(1+x)^k \geq 1 + kx$, for natural numbers $k$ and positive $1+x$. We know the smallest value for $k$ is $2$, so $(1+a+\frac{a^2}{k}-\frac{1}{k})>(1+a-\frac{1}{2})>\frac{1}{2}+a>0$, we thus get by this Bernoulli estimate:
$$a_2^k \geq 1 + k(a+\frac{a^2}{k}-\frac{1}{k})=ak + a^2> 2a +a^2 >a$$
And thus:
$$\sqrt[k]{a}\leq a_{2}\leq a_1$$
Now suppose for some natural number $l$, we have the induction hypothesis:
$$\sqrt[k]{a}\leq a_{l}\leq a_{l-1}$$
Your task, using this, will be to show that now we also have:
$$\sqrt[k]{a}\leq a_{l+1}\leq a_{l}$$
This would complete the proof. Good luck! happy estimating.
