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I have been trying to prove that $e^{x} \geq x^{e}$ for all $x \geq 0$.

This is what I have gotten so far:

$$ f(x) = e^{x} \\ g(x) = x^{e} \\ f'(x) = \frac{d(e^{x})}{dx} = e^{x} \\ g'(x) = \frac{d(x^{e})}{dx} = e x^{e-1}\\ $$ Now for all $x \geq 0$, $f'(x) \geq 0$, so $f(x)$ is always increasing or is constant, and $g'(x) \geq 0$ so $g(x)$ is always increasing or is constant.

Thus across the interval $[0,e]$, the lowest $f(x)$ can be is $f(0) = 1$ and the lowest $g(x)$ can be is $g(0) = 0$. The highest possible value of $f(x)$ is $f(e) = e^{e}$ and highest value of $g(x)$ is $g(e) = e^{e}$.

Thus across the interval $[0,e]$, $f(x) \geq g(x)$.

I've got 2 questions. One, is what I did correct. If not, where did I go wrong and what is the right method. Two, if yes, how does one prove it for the interval $(e,\infty)$.

Stella Biderman
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Neev Parikh
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  • See http://math.stackexchange.com/questions/1619911/why-ex-is-always-greater-than-xe. – S.C.B. Feb 28 '17 at 14:31
  • I don't see this as a duplicate, because this is asking about proof verification primarily. It would be appropriate to link to that answer as a citation for part $2$, but that doesn't answer the question "is this proof I have given correct." – Stella Biderman Feb 28 '17 at 14:42
  • Stella makes a fair point, and in that regard, no, what OP did is not correct. The fact that $f(x)$ and $g(x)$ are both monotonically increasing and the fact that $f(0) > g(0)$ and $f(e) > g(e)$ does not necessarily imply that $f(x) \ge g(x)$ on all of $[0,e]$. This is because $f$ and $g$ can both be always increasing but at varying rates. Maybe $g$ increases faster than $f$ at first, so that $g(x) > f(x)$ on some subinterval, but then $f$ increases faster than $g$ after a certain point. That doesn't happen with these two functions, but my point is that the proof doesn't account for it. –  Feb 28 '17 at 14:52
  • A counterexample to OP's proof idea is $f(x) = x^2+1$ and $g(x) = 3x$. Then $f'(x) = 2x \ge 0$ on, say, $[0, 10]$, and $g'(x) = 3 > 0$. And $f(0) = 1 > 0 = g(0)$ and $f(10) = 101 > 30 = g(10)$. So these two functions meet all of the criteria in OP's proof, but it is not the case that $f(x) \ge g(x)$ on all of $[0,10]$, because, for example, $f(2) = 5$ and $g(2) = 6$. –  Feb 28 '17 at 14:57
  • Thanks for explaining why what I did was incorrect. I'm stumped though. I will check out the other thread. – Neev Parikh Feb 28 '17 at 15:26

1 Answers1

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Your proof is incorrect. Just because both functions are increasing doesn't mean that the one that is ahead always maintains the lead. This is because $g(x)$ could pass $f(x)$ because it starts off increasing sufficiently fast, and then at a later point $f(x)$ could start increasing very rapidly to regain the lead over $g(x)$.

EDIT: a specific counter example has been given in the comments.

Stella Biderman
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  • Can't believe I didn't think of that. Thanks anyway. I'll work on doing it inductively. – Neev Parikh Feb 28 '17 at 15:27
  • I tried to do this with induction. Unfortunately I cannot get past the 'assume true for $x = n$ step'. If you take $e^{n-1}-n^{e-1}$ to be true, and then try proving for $e^{n}-(n+1)^{e-1}$ I cannot get anywhere. – Neev Parikh Feb 28 '17 at 16:29
  • @NeevParikh I had though that $x^{e-1}>(x-1)^e$ held for all $x>2.1$ but actually it holds for all $x>3.1$. I'll see if there's a way to salvage the idea. – Stella Biderman Feb 28 '17 at 16:49