This proof elegantly distills a classic transcendental comparison into a masterclass of pedagogical clarity. It beautifully demonstrates how fundamental inequalities can resolve complex numerical relationships with minimal artifice.
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e^pi vs pi^eAdded:
e^pi vs pi^e, which one's bigger? Well, check this out. I will be using a solution from a viewer from eight years ago. E to the x is greater than 1 + x. And although this is true for all x, I will just write down 4x is greater than zero because the proof for this is easier because if you look at e to the x, the power series suspension for that, we get this thing here. And if you just consider the past dx, if you get rid of all these terms, then you see e to the x has to be greater than 1 plus x. Then we are going to let x equal<unk> / e minus one. And notice this is indeed positive. So we can plug that in. E to the<unk> / e minus one has to be greater than 1 +<unk> / e minus one.
This and that cancel. And then let's multiply both sides by e. This times that add the exponents. We get e to the pi over e is greater than just pi. Then we just have to raise both sides to the e power. And when we multiply exponents, we get e to the pi. And then when we have this, it's just pi to the e. And right here, we still maintain the same inequality because the insides are positive. When we raise that to the positive power, it's an increasing function. That's it.
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