# Inequality on exponents of positive numbers

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.

• Hint: find the unique max of $x^{1/x}$. – Macavity May 26 '16 at 14:43
• could you please explain further Macavity ? How does this link to the problem above ? – Noob101 May 28 '16 at 10:17
• 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. – Macavity Jun 9 '16 at 7:42
• Thanks for pointing this out to me @Macavity . In case you have a different method please let me know . – Noob101 Jun 10 '16 at 9:53

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 $e \le x \le y$.
• @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 $e$. – user21820 Jun 8 '16 at 11:34