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I know that you can prove $e\le y<x$ implies $x^y<y^x$ by the following method:

$$x^y<y^x\iff y \log x<x\log y\iff \frac{\log x}x<\frac{\log y}y,$$

and since $\frac d{dx}\frac{\log x}x=\frac{1-\log x}{x^2}<0$ when $x>e$, $\frac{\log x}x$ is decreasing.

This method doesn't sit well with me, though, since it is a basic number theoretical question ($x$ and $y$ can be restricted to natural numbers). Are there any non-calculus proofs of this statement? (The lower bound $e$ can be relaxed to something larger if necessary, but it has to be a fixed number, not an existence claim.)

P.S. You can now view the Metamath page that arose from this question and its answer: logdivlti

Mario Carneiro
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  • There's this question here http://math.stackexchange.com/questions/9505/xy-yx-for-integers-x-and-y. You might be able to modify the general solution to the equality to obtain what you need. – Ian Coley Mar 10 '14 at 07:01

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Let $t = \frac xy > 1$. Then:

$$\begin{align} \frac{\log x}x = \frac{\log t + \log y}{ty} < \frac{\log y}y &\iff \log t + \log y < t\log y\\ &\iff \log t < (t - 1)\log y\\ &\iff \frac{\log t}{t-1} < \log y\\ &\iff\frac{\log [(t-1) + 1]}{t-1} < \log y. \end{align}$$

Put $z = t-1 > 0$, then we prove: $\frac{\log(z+1)}z < \log y$.

But: $z+1 < e^z \implies \log(z+1) < z \implies \frac{\log(z+1)}z < 1 = \log e \le \log y$. We're done.

Mario Carneiro
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DeepSea
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  • How does this work in the converse situation, i.e. $0<x<y\le e$ implies $\frac{\log x}x<\frac{\log y}y$? The first part works fine, but the transitivity chain flips part-way, so the same argument doesn't work without some modification. – Mario Carneiro Jul 18 '14 at 21:32