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But $f(2)^3$ is not necessarily equal to $f(8)$.

chaotic_iak wrote: But $f(2)^3$ is not necessarily equal to $f(8)$. and $f(25)$ is not necessarily equal to $f(5)f(5)$

$f(27)f(8) \ge f(21)f(10)$ $f(21)f(13) \ge f(27)f(10)$

How do you get $f(27)f(8) \ge f(21)f(10)$ (and the second inequality too)?

Oh whoops forgot "relatively prime" Anyways, aren't both of the inequalities true by the condition "$f(mn) = f(m)f(n)$"? 216>210 and 273>270, and our function is "non-decreasing". Also, all of the pairs are "relatively prime".

Also $f(8).f(19) \ge f(15).f(10)$ and $f(15).f(13)\ge f(19).f(10) $.

Sorry i didn't see that$15$and $10$ are coprime. Though i have found a solution. $f(9).f(8) \ge f(10).f(7) $ and $f(13).f(7) \ge f(10).f(9) $

We want to have 2 inequalities of the form $f(8)f(x) \geq f(y)f(10)$ $f(y)f(13) \geq f(x)f(10)$ Which, when multiplied, gives the desired answer. Thus we require $8x>10y$ and $13y>10x \implies 26y>20x>25y$. Additionally, $(x,8), (x,10), (y,10), (y,13)$ are all relatively prime. These suffice as $f$ is a non-decreasing function. Therefore $x, y$ are odd. Let $n$ be the remainder when $25y$ is divided by $20$. Then, $n+y>20$ for an $x$ to exist. Note $25y \equiv n \pmod {20}$ means that $n=5,15$. Case 1: $n=5 \implies y \equiv 1 \pmod 4 \implies y \geq 17$. Case 2: $n=15 \implies y \equiv 3 \pmod 4 \implies y \geq 7$. 7 satisfies the conditions, so we use $f(8)f(9) \geq f(7)f(10)$ $f(7)f(13) \geq f(9)f(10)$ giving the desired answer. Note that, of course, there are an infinite number of $y$ such that this works. But 7 is the smallest such $y$. Indeed, 21 is the smallest working $y$ for Case 1, as in Sayan's answer.