If the inductor is fully 'corrected' by connecting the capacitor to the inductor, then their total reactance will go to zero, or infinity, depending on whether it's a series or parallel connection. Their reactive power in this case falls to zero.
The output from a lamp connected with them will be the same (assuming ideal, non-lossy L and C) as if it were connected directly to the supply. With zero reactive power, the power factor goes to 1.
That is more light from the lamp than if it was connected to the supply through either the inductor by itself, or the capacitor alone.
With real, lossy, components, there will be some residual R left, mostly from inductor losses, that will dim the lamp slightly with respect to a direct connection to the supply.
However, my assumptions about circuit diagrams may be incorrect, there is another possibility.
If you have a resistor and an inductor in series, then add the capacitor across the supply, to improve the power factor that the supply sees, the brightness of the lamp will remain unchanged, as the voltage on the capacitor, and the voltage across the R+L, is controlled by the low impedance of the supply. What the capacitor will do is to draw current from the supply to cancel out the reactive current drawn by the inductor. When fully corrected for reactive power, the real power drawn by the circuit will be equal to the power used in the lamp. This will be less than had the lamp been connected directly to the supply.