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Let $x>1$, $y>1$ and $x\neq y$ such that $x^y=y^x$. Prove that: $$x^y>e^e$$

We can assume $y=kx$, where $k>1$.

Hence, $x=k^{\frac{1}{k-1}}$ and $y=k^{\frac{k}{k-1}}$ and we need to prove that $f(k)>e^e$, where $f(k)=k^{\frac{k^{\frac{k}{k-1}}}{k-1}}$.

We can show that $f'(k)>0$ for $k>1$ and $\lim\limits_{k\rightarrow1^+}f(k)=e^e$ and we are done.

Is there a nice proof for this inequality? Thank you!

Michael Rozenberg
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  • Related : http://math.stackexchange.com/questions/1705690/prove-that-xy-yx – lab bhattacharjee Mar 20 '16 at 14:44
  • Dear @lab bhattacharjee My problem is absolutely another problem. – Michael Rozenberg Mar 20 '16 at 14:49
  • If you take up your suggestion that you can write $y = kx$ with $k > 1$ a constant, then you can define a function $f(x)$ given as $f(x) = x^y = x^{kx}$. Now take logarithms: $\ln f = k x \ln x$ so that $\frac{1}{f} \frac{df}{dx} = k(\ln x + 1)$. This is zero when $x = 1/e$, so that the (presumably) minimum of $f(x)$ occurs at $x = 1/e$ and $f(x) > (1/e)^{(k/e)}$. Not sure this helps since the turning point occurs for a value $x < 1$? I was expecting to determine the turning point of $f(x)$ at $x= e$. Have I made a mistake in $\frac{df}{dx}$? If the turning point is at $e$ then everything OK. – jim Mar 20 '16 at 19:08
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    Dear @jim I Think your reasoning is wrong because $x$ is a function of $k$. – Michael Rozenberg Mar 21 '16 at 11:02
  • What values of x,y are known to satisfy $x^y=y^x$ for$x\neq y$ ? – NoChance Mar 23 '16 at 20:31
  • @NoChance Yes you are right! Since $2^4=4^2$, we get $16>e^{e}$ without calculator. It's exactly the motivation for my problem. – Michael Rozenberg Mar 23 '16 at 21:39
  • Aha... I see, thx. – NoChance Mar 23 '16 at 21:43
  • When you write "We can show that $f'(k)\gt0$...," are you saying you've done the proof that way, so that the question is to find an alternative, simpler proof of the inequality $x^y\gt e^e$? Or is $f'(k)\gt0$ the inequality for which you'd like a "nice" proof? – Barry Cipra Mar 24 '16 at 13:43
  • You certainly have lim${k \to 1^+}k^{\frac{k}{k-1}} = e$. This follows by considering $z = k^{\frac{k}{k-1}}$ so that $\ln z = \frac{k}{k-1} \ln k$, then lim${k \to 1^+} \ln z =$ lim$_{k \to 1^+} \frac{1 + \ln k}{1} = 1$, i.e. $z=e$. Then I suppose that you can look at $f(k) = z^z$ to write $\ln(f) = z \ln z$ to show that $\ln(f) \to e \ln e$. – jim Mar 24 '16 at 14:50
  • @Barry Cipra I wrote my proof of my problem, but I think that it's very ugly and I am looking for a nice proof. – Michael Rozenberg Mar 24 '16 at 18:59

4 Answers4

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We have to minimize the function $$f(x)=x^{y}\tag{1}$$ where $x$ and $y$ are connected by equation $$x^{y}=y^{x}\tag{2}$$ As usual we need to find derivative $f'(x)$ and this is bit tricky here. First we take logarithm of the equation $(2)$ to get $$y\log x = x\log y$$ and then differentiating with respect to $x$ we get $$\frac{y}{x} + y'\log x = \log y + \frac{xy'}{y}$$ or $$y' = \frac{y}{x}\cdot\frac{y - x\log y}{x - y\log x}\tag{3}$$ From $(1)$ we get $$\log f(x) = y\log x$$ and differentiation gives us $$\frac{f'(x)}{f(x)} = \frac{y}{x} + y'\log x$$ and using equation $(3)$ we get $$\frac{f'(x)}{f(x)} = \frac{y}{x}\left(1 + \frac{y\log x - x\log x\log y}{x - y\log x}\right)$$ and thus $f'(x) = 0$ is possible only when $$\log x\log y = 1\tag{4}$$ so that $y = \exp(1/\log x)$ and $$f(x) = x^{y} = y^{x} = \exp\left(\frac{x}{\log x}\right)$$ It can be easily checked that the function $g(x) = x/\log x$ attains it's minimum value at $x = e$ and therefore $f(x) = e^{g(x)}$ also attains it's minimum value at $x = e$ and therefore the minimum value of $f(x) = x^{y}$ is $e^{e}$.

Paramanand Singh
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I count this as a partial answer: If you keep to the branch $y \le x$ (which given the symmetry of the problem I think is OK), then using the known properties of the Lambert function you have $y = x$ for $0 \le x \le e$. Now consider $x^y$. For $0 \le x \le e$ you have $x^y = x^x$. This solution can be considered "known" in the sense that elementary means can used to deduce its properties, such as general shape. If you know consider the equation $y = -x \frac{W(-\frac{\ln x}{x})}{\ln x}$ and look at $x^y$ then Maple gives the following far from obvious structure (see attached figure).

To complete the answer you would have to show that $\frac{d x^y}{d x} \ge 0$ for $x \ge e$. This looks a fearsome task but it may be possible using only known properties of Lambert's function (for example) $\frac{d W(z)}{d z} = \frac{W(z)}{z (1 + W(z))}$ (from Wiki).

From the figure you seem to have (for the chosen branch) that $x^y > e^e$ for $x > e$. enter image description here

jim
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I think this can be solved by recasting it in the language of Lagrange multipliers. We wish to maximize the function $f(x,y,z) = z$ subject to the constraints $g_1(x,y,z) = x^y - z= 0$ and $g_2(x,y,z) = y^x - z = 0$. By the properties of Lagrange multipliers, $f$ will have a local extremum on the surface specified by these constraints if there exists a point where $$ \nabla f + \lambda_1 \nabla g_1 + \lambda_2 \nabla g_2 = 0. $$ Taking these gradients, we find that we must have \begin{align*} \lambda_1 [y x^{y-1}] + \lambda_2 [(\ln y) y^x] &= 0 \\ \lambda_1 [(\ln x) x^y] + \lambda_2 [x y^{x-1}] &= 0 \\ 1 -\lambda_1 - \lambda_2 &= 0 \end{align*} We can then eliminate $\lambda_2$ from these equations: \begin{align*} \lambda_1 [y x^{y-1}] + (1 - \lambda_1) [(\ln y) y^x] &= 0 \\ \lambda_1 [(\ln x) x^y] + (1 - \lambda_1) [x y^{x-1}] &= 0 \end{align*} Then we can solve each equation for $\lambda_1$ to eliminate it from our system: $$ \lambda_1 = \frac{(\ln y) y^x}{(\ln y) y^x - y x^{y-1}} = \frac{x y^{x-1}}{x y^{x-1} - (\ln x) x^y } $$ Inverting both sides of this equation and simplifying yields $$ \frac{y x^{y-1}}{(\ln y) y^x} = \frac{(\ln x) x^y }{x y^{x-1}}, $$ which further simplifies to $$ (\ln x) (\ln y) = 1. $$ Thus, at the extremum (if it exists), we must have $y = e^{1/\ln x}$ and $z = y^x = e^{x/\ln x}.$ But on the interval $(1, \infty)$, the function $x / \ln x$ has a global minimum at $x = e$. Moreover, since the exponential function is monotonically increasing, the function $z = e^{x/\ln x}$ has a global minimum at $x = e$ as well. Thus, $z \geq e^e$, with equality at the point $x = y = e$.

Michael Seifert
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You certainly have $_{k \to 1^+}k^{\frac{k}{k-1}} = e$. This follows by considering $z = k^{\frac{k}{k-1}}$ so that $\ln z = \frac{k}{k-1} \ln k$, then lim$_{k \to 1^+} \ln z = \frac{1 + \ln k}{1} = 1$ (L'Hopital) i.e. $z = e$ (alternatively, look at the definition of $e = $ lim$_{n \to \infty} (1 + \frac{1}{n})^n$). Then look at $f(k) = z^z$ and write $\ln(f) = z \ln z$ to show lim$_{k \to 1^+} \ln(f) \to e \ln e$, giving lim$_{k \to 1^+} f(k) = e^e$.

jim
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