The question is: for any $n\geq2$, is there always a prime $p$ satifying $\varphi(n)<p\leq n$?
Here $\varphi(n)$ is the Euler totient function.
We know that there is always a prime between $n-O(n^\theta)$ and $n$, where $\theta$ can be $0.525$ (Wiki: Prime gap). Under Riemann hypothesis, one can improve this bound to $O(\sqrt n\log^2n)$. But on the other hand, there are infinite many $n$ such that $\phi(n)\geq n-C\sqrt n$ for some constant $C$ (just choose $n=p(p+k)$ where $p$ and $p+k$ are both prime; for some $k$ these $p$ are infinite). So these upper bound for prime gap don't help.
So can we prove this propsition, or give a counterexample? (or give a evidence to explan why is this hard to prove, maybe?)
(The propsition is equivalent to: if $\varphi(n)>\varphi(k)$ for all $1\leq k<n$, then $n$ is prime)
For prime numbers $n$ the claim is trivial. If $n$ is composite then we have the upper bound $$ \phi(n)<n-\sqrt{n}. $$ If we can show that there is always a prime between $n-\sqrt{n}$ and $n$, for $n\ge 3$, then the claim follows. Unfortunately this is not yet clear, see At least one prime between N and N-(sqrtN).
Reference: Upper bound for Euler's totient function on composite numbers