Assume that $ x\geq y $ By applying AM-GM to the second equality $c^x+d^y = 2^y(cd)^{y/2}$, we get that $2^{y-1} \geq c^{\frac{x-y}{2}}$ so we conclude that $y \geq 1$. Now from the first equality: $a^x+b^y = (a^2+b^2)^x$ we have that: $a^x+b^x \geq (a^2+b^2)^x$ which means that: $1 \leq (\frac{a}{a^2+b^2})^x+(\frac{b}{a^2+b^2})^x \leq \frac{a}{a^2+b^2}+\frac{b}{a^2+b^2}<\frac{a^2}{a^2+b^2}+\frac{b^2}{a^2+b^2}=1$ Contradiction! Conclusion: $x<y$