This seems to be the case for a current drive amp design, but we mainly use voltage drive amp designs, don't we?
Do you understand the difference between an (ideal) voltage source and current source? Ohms law?
I = V/R
An ideal voltage source outputs V regardless of R. If R is high (such as at a resonance peak where the speaker is more efficient) then naturally I drops. If R is lowered then I increases. If R is close to infinity (like an open circuit) then I goes towards 0. If R is close to 0 (like a short) then I goes towards infinity. V stays the same.
In practice, R is not just the speaker's impedance but includes everything outside the amp's feedback loop, like resistance of the speaker terminals, the wires etc. and the voltage across those components splits accordingly.
V = R I
A current source outputs I regardless of R. If R is high (such as at the resonance peak) then V is high. If R is close to infinity (like an open circuit) then V goes towards infinity as well. If R is close to 0 (like a short) then V goes towards 0. I stays the same.
As you increase output resistance (and lower DF towards 0), you effectively turn a voltage source more and more into a current source.
It's very simple to do this with a power/headphone amp, just add a resistor to each channel. But at some point, it becomes impractical. For example, with a 1 kohm output resistor (yielding a DF of 0.008 with 8 ohm load), you'd need your amp to output 126 V unloaded to get just 1V across the 8 ohm load. (Btw, if the load drops to 2 ohms then voltage drops to 0.25 V, which is like a 12 dB difference).
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Analog audio signals are typically transmitted as voltages. Power amps may be current sources in integrated and special designs, but make little sense in general use cases. Most off-the-shelf power amps are voltage sources. Most passive speakers are designed to be driven by those.
Headphone outs historically had like 120 ohm output impedance, but dedicated headphone amps nowadays are typically very low output impedance as well.