The problem is trivial if at least one of $x$ or $y$ is greater than $1$. So all we need is to proof that $x^y+y^x > 1 \ \forall x,y \in (0,1)$.
2026-05-17 12:17:38.1779020258
Proof that $x^y + y^x > 1 \ \forall x,y > 0 $
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By Bernoulli $$\frac{1}{x^y}=\left(1+\frac{1}{x}-1\right)^y\leq1+y\left(\frac{1}{x}-1\right)=\frac{x+y-xy}{x}.$$ Thus, $$x^y\geq\frac{x}{x+y-xy}>\frac{x}{x+y}$$ and we are done.