Prove the equation has unique class of solutions

Question

Find the solutions of equation: $$ x^y + y^x = 1 + xy \quad x,y \in \mathbb{R} \quad x,y >0 $$

My quest

First, $x=1$ or $y=1$ gives us obvious solutions, so let's suppose $x \not =1$ and $y \not= 1$.

Using Bernoulli's inequality I was able to show the inequality: $$ x^y + y^x > 1 + xy \quad \forall x,y>0 \quad and \quad (x-1)(y-1) \gt 0 $$ Therefore there is no solution with $x,y\gt1$ or $x,y\lt1$.

My goal is to show that there is no solution having $x \not =1$ and $y \not= 1$ and the only case I was not able to cover is $(x-1)(y-1)<0$.

Any help will be appreciated.

Answer 1

I've got it:

Suppose $x > 1, y < 1$. Then, using Bernoulli's inequality: $$ x^y = (1 + (x - 1))^y < 1 + y(x -1) $$ Therefore $$ x^y + y^x < 1 + y(x -1) + y^x= 1 + xy -y + y^x < 1 + xy $$ because: $y^x<y$

Answer 2

By the fundamental theorem on solutions to non linear equations, there are exactly n - 1 solutions, where n is the number of variables

Answer 3

you have given no source and no indication of your background. The conclusion that the solution set is the lines (in the first quadrant only) $y=1$ and $x=1$ seems quite likely given these plots; if you know multivariable calculus this gives some directions to pursue.