Will the charging current decrease due to a higher input resistance?
Yes but, but, but ... . The basic method is correct as far as it goes but there are more factors to consider than are covered by that formula. 

The figure is essentially correct 


for an ideal 5V supply   
and a battery whose Ah capacity is numerically much larger than Imax_A,   
if you wish to charge at true constant current in the CC phase, 
as long as your assumed resistances are correct.


If you can tolerate a decreasing current during CC then Icharge can be greater when Vbattery is &lt; 4.1V, provided that you do not exceed the maximum available current from the "USB" source. eg a 1000 mAh battery may allow C/1 max charge rate = 1A. Your formula may give 600 mA in a specific case, but when Vbat is below 4.1V Imax will increase. Your formula gives Reffective = V/I = 0.6 / 600 mA = 1 Ohm. At Vbattery = 3.7V the formual yields 1A and for Vbattery &lt; 3.7V you will need to current limit in CC mode t present the C/1 rate being exceeded. 

One assumption that you made but did not explicitly state is that V_USB === 5V.
Actual is ~= 5V. You may look at the spec to see what the range of "~+" is in theory. In practice it can be somewhat higher than 5V and may be less. 

Note that while that gives the maximum theoretical current for a 5V USB supply, the available current may be restricted to less than this theoretically and/or actually by the supply itself.

Also, maximum battery current must not be more than the manufacturer allows - usually C/1 but eg C/2 in some cases and I've seen 2C allowed for some "special cells.