You are simulating a circuit with extremely crappy power factor. If the device in question really works at 30VDC, that's a clear sign the diode conduction duty cycle is much longer, and the ratio of peak to average current more benign.
Besides, given that power loss in diode isn't I^2*R but more like I*Vf, diodes handle peak currents quite well; in other words, diodes are always rated for average current (look at any diode datasheet) because that defines the die heating, not RMS current like with MOSFETs or capacitor ESR. Average current for peaky or more steady current flow is the same, so is heat generated in the diode. So as long as the peak current is within specifications, it's OK.
But only using two diodes out of the diode bridge really doubles the average current, this is not insignificant.
You are right though that peaky current affects the capacitor power dissipation. Obviously the cap must be sized to handle it. Going for DC completely removes the capacitor ESR loss. But the capacitor ESR loss in such circuits often isn't that much to begin with.
Inrush current obviously is linearly related to the input voltage (AC or DC). Going lower lowers the inrush, but unless the OP plans to abuse the light by disco blinking it, it doesn't matter because it has to be designed for some quite high number of on/off cycles (often stated in the specs, even) at full rated voltage.
To recap:
* By going for smoother (DC) current instead of peaky AC, diode power dissipation lowers only slightly (due to the logarithmic nature of Vf ~ I), not by a lot as you seem to expect
* Diodes have to be rated for the full peak current during inrush at full line peak voltage, otherwise the devices would go up in smoke all the time
* Diode thermal rating is designed for the average current, not peak current
* Feeding DC to diode bridge doubles the average current two diodes see
But by all means plot the diode power and calculate the integral in your Spice simulation, and compare to the DC input. I expect you may see some 10% difference.