Team Selection Test for JBMO

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 $x<y$.