Standard incandescent light bulbs, which is what most enlarger bulbs are, pass current through a piece of tungsten wire. The wire gets really hot, and glows. The "resistance" of the wire is much lower when it's cold, which means it will conduct much more current (amps) when you first turn it on. It heats up very quickly, increasing its resistance along the way, and quickly arrives at an equilibrium where it gives off constant light. This process takes maybe 0.05sec for most bulbs. (You can think of resistance as the diameter of a pipe. The wider the pipe, the lower the resistance, so more water can flow through it. A lightbulb is like a dynamic pipe where the diameter decreases rapidly as it heats up.)
Most enlargers bulbs are powered directly from house AC (alternating current). The voltage in your walls is AC - the exact voltage goes positive and negative relative to "neutral", but the "rms" value is 120V (or 110V, or 230V, depending where you are), and that's how we refer to it. [RMS is a method of mathematical power averaging. It doesn't matter for this discussion.] Because it's AC and varies with time, there are times when the voltage is very small, and times when it is large, it just depends where in the cycle you happen to press the power switch. For 120VAC in North America, the peaks are actually about 170V. As Gerald pointed out, bulbs typically fail if the switch is turned on when the voltage is near one of its peaks, which happens 100 or 120 times per second (50Hz vs 60Hz). If you get the timing just right, and we've all had this happen, you'll get an intensely bright flash as the bulb's instanteous power is much much higher than is rating, followed by darkness.
There's also an aging effect related to operating voltage. The higher the voltage, the shorter the life. The reason is that tungsten molecules slowly "boil" off the filament wire when the bulb is on and end up on the cooler bulb envelope (the glass). This causes the filament to get thinner as it ages. A thinner wire will be more prone to fail during turn-on if you happen to hit one of the line-cycle peaks. That's why new bulbs rarely fail on turn-on, while older bulbs are more likely to do so.
Using an in-line resistor to reduce the bulb's operating voltage will certainly extend its life, but will reduce its brightness and increase printing time. It also reduces the likelihood of popping the bulb when you turn it on because it will limit the current. A side-effect of this current-limiting is an increase in turn-on time. I don't know if this is a practical issue for enlarging, I've never tested it, but I would guess it shouldn't be too big an effect, maybe a tenth of a second or a bit more.
An inductor should limit turn-on current without affecting brightness. Because of this it won't extend bulb life. But inductors always have internal resistance as well, so in practice you'd need an inductor built with heavy-gauge wire if you didn't want to affect brightness. And because of the way inductors work, you need to think about what happens when you turn the light off, there can be a big voltage spike.
A diode will cut the bulb power in half, extending its life considerably, and reducing its brightness quite a bit. It doesn't do anything to prevent bulb failure at turn-on because it has essentially no effect during half of each AC line cycle (while completely blocking the other half-cycle).
The best solution, and something I've been meaning to get around to building for myself, is what's called a "zero voltage switch". Conceptually, it watches the AC voltage and only turns on the bulb when the voltage is small - it's not exactly zero, but is small enough that it doesn't matter. This protects the bulb against turn-on problems with no sacrifice in light output. This is more complex than just a resistor or inductor, but with today's electronic components it's fairly easy to do. But since my enlarger uses a cold light it's not something I need.