Scope watching voltage at inverter would tell you something. Precharge resistor serves as current sense resistor.
Clicking with repeated power up attempts is dissipating power somewhere.
Some switching power supplies commit Hari-Kari under brownout conditions. Inverter may have those to power internal circuitry.
a 10 ohm resistor would power up faster.
My suggestion (for BMS without precharge system built in) is divide battery voltage (e.g. 50V) by maximum current spec (something like 100A or 200A), which gives a resistor value 0.5 ohm or 0.25 ohm. That means 50V into 0.5 ohm would draw 100A max. Use that for resistor.
A length of 12 awg or 10 awg Romex, with white and black wire joined by wire nut at one end, would be good. Some 10's to 100's of feet, a spool, would be 0.5 ohm, and power dissipation would be quite good. It can carry a massive current like 100A for a short period of time 1000A, sufficient to precharge capacitors and then power inverter. Connect this precharge resistor then close main switch.
I think inverter has 0.1F to 1.0F of capacitor (some would be less.) T = RC = 0.5 x 1.0 = 1/2 second to charge 1.0 F to within 1/e of battery voltage or about 2/3 of the way there. Six time constants, 3 seconds, gets within 2^6/3^6 = 0.09 of charged, 91% of the way there. (more if smaller capacitor)
Your 100 ohm resistor and 0.1 F would have 10 second time constant, a minute to reach 90%! And that's if no significant DC draw.
10 ohm resistor would have 1 second time constant, so 6 seconds to charge the smaller assumed 0.1 F capacitor. 60 second for 1.0 uF. 0.6 seconds for 0.01 F (10 mF, 10,000 uF)
If BMS has precharge circuit, that is likely decent performing for some range of inverter input capacitor. But not if inverter starts drawing current. In that case, having BMS precharge a supercap that is 10F or larger, then using Romex 0.5 ohm "resistor" to precharge inverter might work.
Bottom line, it take a low value resistor or inordinately long time with higher value resistor to charge typical large inverter input capacitor bank. And with any discrete resistor, it can't power inverter through the resistor, needs to be turned completely on.