Classic Algebra Question II

$\frac{xe^{y-x}+ye^{x-y}}{2} \geq \sqrt{(xe^{y-x})(ye^{x-y})} = \sqrt{xy}$ $\begin{aligned} \frac{xe^{y-x}+ye^{x-y}}{2} &= \sqrt{xy} \\ \iff xe^{y-x} &= ye^{x-y} \\ \frac{x}{y} &= e^{2(x-y)} \\ \end{aligned}$

Since $e^{2(x-y)}$ is a rational number, then the value of $x-y$ will be equal to $\boxed 0$ .

$\begin{aligned} \frac {xe^{y-x}+ye^{x-y}}2 & = \sqrt{xy} \\ \frac x{e^{x-y}} + ye^{x-y} & = 2\sqrt{xy} & \small \color{#3D99F6} \text{Squaring both sides} \\ \frac {x^2}{e^{2(x-y)}} + 2xy + y^2e^{2(x-y)} & = 4xy \\ \frac {x^2}{e^{2(x-y)}} - 2xy + y^2e^{2(x-y)} & = 0 \\ \left(\frac x{e^{x-y}} - ye^{x-y}\right)^2 & = 0 \\ \implies \frac x{e^{x-y}} & = ye^{x-y} \\ \frac xy & = e^{2(x-y)} \end{aligned}$

Note that the LHS is rational. For the RHS to be rational, there is only one solution, which is $x-y=\boxed 0$ .