A follow-up question on an arithmetic function satisfying a certain inequality In the MSE question here, I asked whether the inequality:
$$\frac{a}{b} + \frac{b}{a} < \frac{f(a)}{f(b)} + \frac{f(b)}{f(a)}$$
would imply $a < b$ (where $f(x) \in \mathbb{N}$ is a function for $x \in \mathbb{N}$), if $\gcd(a, b) = 1$.  For the example I'm considering, I actually have $f(1) = 1$ and $f$ multiplicative (i.e., $f(mn) = f(m)f(n)$ whenever $\gcd(m, n) = 1$).  (In fact, I do have $f(x) = \sigma_{1}(x)$, where $\sigma_{1}$ is the sum-of-divisors function.)
It turned out (from the subsequent answer by André Nicolas and a comment by Marc van Leeuwen) that the first inequality cannot force the second, and that the first cannot imply the second unless the (first?) inequality is absurd (i.e., if it holds for no pair $(a,b)$ at all; ex falso sequitur quodlibet).
Now my follow-up question is this:  Will the statements from the previous paragraph still be true if we have the added condition that
$$\frac{f(a)}{f(b)} < \frac{a}{b}?$$
Edit [Feb 19 2013] - From a subsequent comment by Ivan Loh, it turns out that if
$$\frac{f(a)}{f(b)} < \frac{b}{a}$$
is also true, then the conclusion that I require will not necessarily follow, granted $a$ and $b$ are prime powers.
What will happen if $a = q^k$ where $q$ is prime and $b$ is a (fairly large) composite number?
 A: The following is a consolidation of my comments into an answer:
A multiplicative function is uniquely determined by its values at the prime powers. Consider any multiplicative function $f: \mathbb{N} \rightarrow \mathbb{N}$ such that for some prime powers $p^m, q^n$ with $p^m<q^n$, we have $\frac{f(q^n)}{f(p^m)}<\frac{p^m}{q^n}$.
Then we can take $a=q^n, b=p^m$, and we get 
$$\frac{q^n}{p^m}+\frac{p^m}{q^n}<\frac{f(q^n)}{f(p^m)}+\frac{f(p^m)}{f(q^n)} \; \text{and} \; \frac{f(q^n)}{f(p^m)}<\frac{p^m}{q^n}<\frac{q^n}{p^m}$$
but $a>b$, giving a counterexample to the claim.
For the explicit example $\sigma_1{(x)}$, trying small cases gives the counter example $a=7, b=6$, since 
$$\frac{7}{6}+\frac{6}{7}<\frac{\sigma_1{(7)}}{\sigma_1{(6)}}+\frac{\sigma_1{(6)}}{\sigma_1{(7)}} \; \text{and} \; \frac{\sigma_1{(7)}}{\sigma_1{(6)}}<\frac{6}{7}<\frac{7}{6}$$
but $7>6$, giving a counterexample to the claim.
In general, all counterexamples must satisfy 3 conditions:
$$a>b, \frac{a}{b}+\frac{b}{a}<\frac{f(a)}{f(b)}+\frac{f(b)}{f(a)}, \frac{f(a)}{f(b)}<\frac{a}{b}$$
We must necessarily have $$a>b, \frac{f(a)}{f(b)}<\frac{b}{a}<\frac{a}{b}$$
This is equivalent to $af(a)<bf(b)$ for $a>b$. 
Thus your claim is correct if and only if $xf(x)$ is a non-decreasing function of $x$.
