Let x and y be positive numbers . Which of the following always implies $x^y \geq y^x $

a.) $x\leq e\leq y$

b.)$y\leq e \leq x$

c.)$x\leq y \leq e$ or $e\leq y \leq x$

d.)$ y\leq x \leq e$ or $e\leq x \leq y $

My attempt : To solve the following inequality I tried taking log on both sides , however I could could not progress much farther than that. I tried assuming two cases after that one where both x and y are greater than e and one where both are less than e.However I could not proceed further. Please help.

  • $\begingroup$ Hint: find the unique max of $x^{1/x}$. $\endgroup$
    – Macavity
    Commented May 26, 2016 at 14:43
  • $\begingroup$ could you please explain further Macavity ? How does this link to the problem above ? $\endgroup$
    – Noob101
    Commented May 28, 2016 at 10:17
  • $\begingroup$ You need to prefix @ to the user name to draw their attention, otherwise your comment may never reach them. In this case I did not get your message till browsing through after a long while. I see you have an answer that explains - in case you need more explanation let me know. $\endgroup$
    – Macavity
    Commented Jun 9, 2016 at 7:42
  • $\begingroup$ Thanks for pointing this out to me @Macavity . In case you have a different method please let me know . $\endgroup$
    – Noob101
    Commented Jun 10, 2016 at 9:53

1 Answer 1


Take positive reals $x,y$. Then $x^y \ge y^x$ iff $y \ln(x) \ge x \ln(y)$ iff $\frac{\ln(x)}{x} \ge \frac{\ln(y)}{y}$.

Then you can find the intervals on which the function $x \mapsto \frac{\ln(x)}{x}$ on the reals is increasing or decreasing, which is easy by using differentiation and Rolle's theorem.

You will find that is strictly increasing on $(0,e]$ and strictly decreasing on $[e,\infty)$, which by definition means that for any reals $x,y$, if $0 < y \le x \le e$ then $\frac{\ln(x)}{x} \ge \frac{\ln(y)}{y}$ and hence (by the chain of equivalences earlier) we get $x^y \ge y^x$, and similarly if $e \le x \le y$.

  • $\begingroup$ I could not follow how finding the global maximum of the function would be of any help.Why can't it be the case that even though the function is not at its peak , but it is still bigger than the RHS ? So I found out that the maximum of the solution occurs at e . But could not link it to any of the options Would you please complete the answer , by showing how this links to the options mentioned above. $\endgroup$
    – Noob101
    Commented Jun 8, 2016 at 10:50
  • $\begingroup$ @SuryakantShrivastava: I've added the remaining steps. The maximum point at $e$ tells you that you can't tell if $x,y$ fall on opposite sides of $e$. It's true that the maximum point doesn't tell you the whole story, so you still need the fact that it is monotonic on either side of $e$. $\endgroup$
    – user21820
    Commented Jun 8, 2016 at 11:34

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