Problem 854

On the contrary suppose that $x - y = t \ge 0$. From $c^x + d^y = 2^y (cd)^{y/2}$ we get
$$\left(\frac{c}{d}\right)^y \cdot c^t + 1 = \left(2\sqrt{\frac{c}{d}}\right)^y$$
Since $t \ge 0$ and $c > 1$ we have $c^t \ge 1$ and
$$\left(2\sqrt{\frac{c}{d}}\right)^y \ge \left(\frac{c}{d}\right)^y + 1$$
Let $\sqrt{\frac{c}{d}} = u$. Since
$$2^y \ge u^y + u^{-y} \ge 2$$
we get $y \ge 1$ and $x = y + t \ge 1$. Therefore,
$$a^x + b^y = (a^2 + b^2)^x \ge a^{2x} + b^{2x}$$
But $a^x < a^{2x}$ and $b^y \le b^x < b^{2x}$. Contradiction shows that $x < y$.