Is the following statement true?


Let $N$ be sufficiently large, and choose $t$ uniformly randomly in $\{1,2,\ldots,N\}$. Then
$$\Pr[\gcd(t, N)>N^{3/4}] < N^{-1/16}.$$


This is the "dual" question to
[Lower bounding the probability that $\gcd(t,N)≤B$, for a random $t$ and fixed (large) $N$](https://mathoverflow.net/questions/212741/lower-bounding-the-probability-that-gcdt-n%e2%89%a4b-for-a-random-t-and-fixed-l) which asks for a bound on the probability that $\gcd$ is small.
In the linked post it is pointed out that


$$\lim\_{N\to \infty} \Pr[\gcd(t, N)>N^{3/4}] = 0.$$


However, the quantitative bound given by applying Markov's Inequality to the expectation computed by Terry Tao is too weak for my use case: Markov's inequality gives a probability of something like $1/\log N$.


I think I've proved the statement for $N$ a prime power.
If there is a counterexample to this statement, e.g., primorials or something, that would be good to know as well.



Probability of large gcd
The statement is true in the stronger form that
$$\Pr[\gcd(t, N)>N^{3/4}] < N^{-1/2}.$$
Indeed, the probability that $\gcd(t,N)$ equals a given $k\mid N$ is at most $1/k$. For $k>N^{3/4}$, this is less than $N^{-3/4}$, hence the probability in question is less than $d^\*(N)N^{-3/4}$, where $d^\*(N)$ is the number of divisors of $N$ exceeding $N^{3/4}$. As $d^\*(N)<N^{1/4}$, the claim follows.


