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Given only the expressions $x^y$ and $y^x$ and no additional information except $x\neq y$ (and the meta-knowledge that the problem was presented in the context of induction), is it possible to determine which is bigger?

If there is no clear solution, am I supposed to glean some additional insight from this problem?

Martin Sleziak
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Vegard
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    See, $2^3<3^2$, but $3^2>2^3$. So it can be either way. BTW, $x^y$ is not an equation, but who cares. – Ivan Neretin Feb 15 '16 at 12:56
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    This problem is completely symmetric in $x$ and $y$ - how would we be able to tell which one is larger? If $x^y>y^x$, then we can simply swap $x$ and $y$ to obtain a different result –  Feb 15 '16 at 12:57
  • I know they are not equations, but I didn't know what else to call them... units? Expressions? – Vegard Feb 15 '16 at 12:57
  • Can you reprint the problem exactly as it appeared? As it stands there's really no way to say anything useful. – lulu Feb 15 '16 at 12:57
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    This has been discussed in many variations at MSE, e.g., see here for $x=e,y=\pi$ ( I suppose it is about something like this). – Dietrich Burde Feb 15 '16 at 12:58
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    The one which has base which is in some sense "closest" to $e$ is the biggest one. – Arthur Feb 15 '16 at 12:59
  • @lulu This is exactly the problem as it was presented. I don't think anyone was meant to find an answer, it was presented as an "over-the-weekend" nut by a prof I had years ago. But I never figured out what he meant us to gain by pondering it, so I thought I'd ask MSE. – Vegard Feb 15 '16 at 13:01
  • You are not given $x^y$ and $y^x$, since if you were then just look at them to decide which is larger. Do you mean "Given distinct $x$ and $y$" instead? – KCd Feb 15 '16 at 13:01
  • @Vegard: The correct term would be "expressions," or perhaps "formulas." – Cameron Buie Feb 15 '16 at 13:02
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    See http://mathforum.org/library/drmath/view/70271.html for additional discussion. – S.C.B. Feb 15 '16 at 13:05
  • @DietrichBurde, my point was that the question is badly written. You could ask, as the question you linked to does, if it can be determined which of $a^b$ or $b^a$ is larger if you know ("are given") $a$ and $b$ without having to directly calculate $a^b$ and $b^a$. What I find poor form is phrasing this as "Given $a^b$ and $b^a$..." when "Given $a$ and $b$..." is meant. – KCd Feb 15 '16 at 13:10
  • @KCd: I'm sorry if the phrasing was poor. In this problem, we don't get to know any of the variables or the expressions. – Vegard Feb 15 '16 at 13:11
  • @MXYMXY That looks very promising, thanks a lot. In combination with the other MSE link and Arthur's statement, I think I am beginning to see where this is going. – Vegard Feb 15 '16 at 13:14
  • This seems related, too: Fastest way to check if $x^y > y^x$? – Martin Sleziak Feb 18 '16 at 17:56

3 Answers3

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For positive integers, the nontrivial case is when $x \neq y$ and both are at least $2$.

Then one is asking which is a more efficient way to partition $xy$ into several (equal) numbers for the purposes of getting a large product: multiply $x$ copies of $y$, or $y$ copies of $x$. It turns out that having a larger exponent is more important. If $2 \leq x < y$, then $x^y$ should exceed $y^x$, but this neat and simple pattern is rudely disrupted by brute calculation that $2^4 = 4^2$ and $2^3 < 3^2$.

As in the real-valued problem usually solved with calculus, the key is the function $F(t) = t^{1/t}$. The inequalities $x^y > y^x$ and $F(x) > F(y)$ are equivalent, reducing the problem to a comparison of values of $F$ at different points. For positive integers larger than $1$, the maximum value is $F(3)$, and $F( )$ is decreasing starting at $3$, but $F(2)=F(4)$. This implies that the exceptions observed above are the only ones to the pattern that it is more efficient to increase the exponent than the base.

For an induction proof without calculus it looks like the thing to prove is $F(n) > F(n+1)$ for $n \geq 3$, which is the same as $n^{n+1} > (n+1)^n$, or $n > (1 + \frac{1}{n})^n$. It is known that $(1+\frac{1}{n})^n$ increases toward $e$ which is less than $3$, which settles the problem. Without relying on that one can replace $n-1$ of the factors in $(1+1/n)^n$ by a telescoping product of terms $(n + 1-i)/(n-i)$, for $i=0$ to $(n-2)$, which leaves an inequality similar to $n > (n/2)$ that can be checked easily and implies the one on $F(n)$.

zyx
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If they are positive integers, the answer is easy:

I will assume that $x < y$.

If $3 \le x$ then $x^y > y^x$.

If $x = 2$ then $2^y < y^2$ for $y=3$; $2^y=y^2$ for $y=4$; $2^y > y^2$ for $y > 4$.

If $x$ and $y$ are reals which exceed $1$, then, since $x^{1/x}$ has a maximum at $x=e$ and is increasing before and increasing after, if $x \ge e$ then $x^y > y^x$.

If $x < e < y$, interesting things can happen. I recall a result of mine from probably 40 or more years ago of the form "if $1 < x < e < y$ and $xy > e^2$ then $x^y < y^x$". The last inequality may go the other way. I'll see if I can find or prove it.

marty cohen
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Actually it's easy for x and y positive. Assume that we've labeled x and y so that y > x. Compute t = y/x. Then

u = t^(1/(t-1)).

If x > u then x^y > y^x. If x < u the reverse is true.

Example y = 3, x = 2. Then t = 3/2, u = 9/4. x < u so x^y < y^x, i.e. 2^3 < 3^2.

Oscar Lanzi
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