Step-by-step Solution:

1. Relate Power Rating to Operating Resistance: Incandescent bulbs are designed to be connected in parallel to a constant voltage source, V (e.g., the mains supply). The power rating (P) of a bulb corresponds to the power it dissipates at its normal operating temperature, when it is glowing. The relationship between power (P), voltage (V), and resistance (R) is given by: $P = \frac{V^2}{R_{op}}$ where $R_{op}$ is the resistance of the bulb's filament at its high operating temperature.

   From this formula, we can express the operating resistance in terms of power and voltage: $R_{op} = \frac{V^2}{P}$ This equation shows that for a constant voltage V, the operating resistance of a bulb is inversely proportional to its power rating, i.e., $R_{op} \propto \frac{1}{P}$.

2. Compare the Operating Resistances: Let's denote the operating resistances of the 100 W, 60 W, and 40 W bulbs as $R'_{100}$, $R'_{60}$, and $R'_{40}$ respectively. Using the relationship from Step 1:

   • $R'_{100} = \frac{V^2}{100}$
   • $R'_{60} = \frac{V^2}{60}$
   • $R'_{40} = \frac{V^2}{40}$

   Since the denominators are in the order $100 > 60 > 40$, the values of the resistances will be in the reverse order: $R'_{100} < R'_{60} < R'_{40}$ So, at their operating temperatures, the 100 W bulb has the lowest resistance, and the 40 W bulb has the highest resistance.

3. Relate Room Temperature Resistance to Operating Resistance: The question states that the resistance of the filament increases with temperature. This is a property of metallic conductors. The resistances measured at room temperature ($R_{100}, R_{60}, R_{40}$) will be lower than their respective operating resistances ($R'_{100}, R'_{60}, R'_{40}$).

   Furthermore, a higher power bulb dissipates more energy and thus operates at a higher temperature. Therefore, the operating temperatures are ordered as: $T_{op, 100} > T_{op, 60} > T_{op, 40}$ The increase in resistance from room temperature to operating temperature is larger for bulbs with higher operating temperatures. This means the ratio $R'_{P}/R_{P}$ is largest for the 100 W bulb and smallest for the 40 W bulb.

4. Compare the Room Temperature Resistances: We established the order of operating resistances as $R'_{100} < R'_{60} < R'_{40}$. Since the filaments are made of the same material, and they are designed to work under the same voltage, it is a reasonable assumption that the order of resistances will be the same at room temperature. The effect of temperature change reinforces this conclusion. The 100 W bulb, having the lowest operating resistance, also experiences the largest temperature increase, which means its room temperature resistance must be significantly lower than its operating resistance. Conversely, the 40 W bulb has the highest operating resistance and the smallest temperature increase. Both factors ensure that the order of resistances is maintained.

   Therefore, the relationship between the resistances at room temperature is: $R_{100} < R_{60} < R_{40}$

5. Evaluate the Options: We need to find the option that is equivalent to the inequality $R_{100} < R_{60} < R_{40}$.

   • Option A: $\frac{1}{R_{100}} = \frac{1}{R_{40}} + \frac{1}{R_{60}}$ - This is a specific formula for parallel resistors and is not a general inequality.
   • Option B: $R_{100} = R_{40} + R_{60}$ - This implies $R_{100}$ is the largest, which contradicts our finding.
   • Option C: $R_{100} > R_{60} > R_{40}$ - This is the reverse of our derived inequality.
   • Option D: $\frac{1}{R_{100}} > \frac{1}{R_{60}} > \frac{1}{R_{40}}$ - When we take the reciprocal of a series of positive numbers, the inequality signs are reversed. Taking the reciprocal of our derived inequality $R_{100} < R_{60} < R_{40}$ gives: $\frac{1}{R_{100}} > \frac{1}{R_{60}} > \frac{1}{R_{40}}$ This matches Option D.