In some car racing games or sports, you have a sort of power-boost you can use for a short amount of time. This question will consider a driver with two different speed boosts available; should they use both at the same time or one after the other?
Suppose that the car has power P normally, and has two ways to briefly increase this. While using the first power-boost, the power is increased to P + P1 for T seconds. While using the second power-boost, the power is increased to P + P2 for T seconds. If both power-boosts are used simultaneously, the power is increased to P + P1 + P2 for T seconds. The velocity of the car in any case is the cube root of its power.
Should the driver use one power-boost and then immediately use the other, or would it be better to use both power-boosts together, and then drive normally for T seconds– which of these options makes the car go further over the course of 2T seconds?
*This isn’t dimensionally consistent; there should be some sort of pre-factor constant, but it turns out that the decision in this question doesn’t depend on the constant. The idea behind the cube root is that the drag on the car is roughly proportional to the square of the velocity, and this should balance the force from the engine (ignoring any time it takes to accelerate up to the new top speed). Power is force times velocity, so will be roughly proportional to the cube of the velocity.
I got this equation for distance travelled using the first way: $\sqrt[3]{P+P1}*T + \sqrt[3]{P+P2}*T$
and for the second way: $\sqrt[3]{P+P1+P2}*T + \sqrt[3]{P}*T$
How should I continue?
What you want to show is that $\sqrt[3]{P+P1+P2}+\sqrt[3]P \lt \sqrt[3]{P+P1}+\sqrt[3]{P+P2}$ This comes from the fact that the cube root grows less than linearly. You can get rid of the $P$ by dividing through by its cube root. A simple example is if the boosts are the same size as $P$, where this becomes $1+\sqrt[3]3 \lt 2\sqrt[3]2$ You can prove this by cubing both sides. The terms where you cube the individual terms cancel and you get more cross terms on the right.