Let $p(n)$ is the biggest prime divisor of $n$. Prove that, exist infinite $n \in N$ satify : $p(n) Let $p(n)$ denote the biggest prime factor of $n$.
Prove that, there are infinite $n \in N$ satisfy : $p(n) <p(n+1)<p(n+2)$
My idea is finding some special consequences like $a_n=2^{2^m+1}$ and $n=a_n$
$p(n+1) \geq 3>2=p(n)$ and $p(n+2)=2(2^{2^m}+1)$ where $2^{2^m}+1$ is Fermat number
If there are infinite Fermat number the problem is solved. (but this statement is more difficult than the problem),
Thank you for helping, Finally I solved it
With same previous idea . Let p are odd prime 
Exist $k_0 = \inf \{k \in N| P(p^{2^k}+1) > p\} ; k_0 < \infty .$
($P(n)$ same meaning with $p(n)$)
By using Lemme : $P(p^{2^{k_0}}+1) \equiv 1 (\mod 2^{k_0+1})$
And then , we prove : $P(p^{2^{k_0}}-1)<P(p^{2^{k_0}})< P(p^{2^{k_0}}+1).$
 A: From page 320 of "On the largest prime factors of $n$ and $n+1$" by Paul Erdos and Carl Pomerance (Aequationes Mathematicae 17, 1978, pp. 311-321):

Suppose now $p$ is an odd prime and
$$k_0=\inf\{k:P(p^{2^k}+1)\gt p\}$$
(note that $P(p^{2^{k_0}}+1)\equiv 1$ mod$(2^{k_0+1})$, so
  $k_0\lt\infty$).  Then
$$P(p^{2^{k_0}}-1)\lt P(p^{2^{k_0}})\lt P(p^{2^{k_0}}+1)$$

Remark:  They go on to say:

On the other hand, we cannot find infinitely many $n$ for which
$$P(n)\gt P(n+1)\gt P(n+2)$$
but perhaps we overlook a simple proof.

A: By Schinzel's hypothesis there exists intfinitly many natural numbers $k$ such that each of the following is prime: 
$$ \ \ \ \ \quad \quad q(k)=3k+2 \quad < \quad r(k)=4k+3 \quad < \quad s(k)=6k+5 \ \ ;\qquad$$
then let $n_k=12k+8$, note that: 
$$ \quad \quad \quad \ \ \ \quad \quad n=4q(k) ,\quad \quad  \quad n+1=3r(k) ,\quad \quad \ \ \ \   n+2=2s(k) \ \ \qquad 
\Longrightarrow
\\ 
p(n) \quad \quad < \quad \quad p(n+1) \quad \quad < \quad \quad p(n+2) .$$

This proof is again built on an open problem;
but notice that the general belief about Fermat numbers is:
$$f_n \ \  \text{is composite for all} \ \ n \geq 5.$$ 
A: I will just exemplify the idea of @Famke: 
If $p$, $\frac{3p+1}{2}$ and $3p+2$ are all prime then for the numbers 
$$3p, 3p+1, 3p+2$$ the largest prime factors are, respecively, those above primes. Now, among the first $10^N$ primes there are , respectively, $39$, $249$, $1546$ ( for $N = 3,4,5$ )  primes $p$ satisfying the condition above. So indeed, testing strongly suggests that the Schinzel conjecture is true, and so we get lots of examples. 
Concrete example : $p= p_{993}= 7867$. Then we get 
$$3\cdot 7867=23601, \ \ 2\cdot 11801= 23602,\ \  23603$$
