a two-variable cyclic power inequality $x^y+y^x>1$ intractable by standard calculus techniques If $x$ is in the open interval (0,1) and so is $y$, prove that $$x^y+y^x>1$$
A direct two-variable application of maxima and minima seems difficult.

Let's suppose(if it's not true, change the place of $x,y$ in the following arguments) $$\dfrac{x^{y-1}}{y^{x-1}} \geq 1$$
Then we have $ (x^y + y^x)^\frac{1}{x} = (y^x (1+\dfrac{x^y}{y^x}))^\frac{1}{x} = y (1+\dfrac{x^y}{y^x})^\frac{1}{x} $
Then since $\frac{1}{x} > 1$, we have $$(x^y + y^x)^\frac{1}{x} = y (1+\dfrac{x^y}{y^x})^\frac{1}{x} \ge y(1+\dfrac{x^y}{y^xx}) = y(1+\dfrac{x^{y-1}}{y^x}) = y + \dfrac{x^{y-1}}{y^{x-1}} \geq y+1 > 1$$
since $\frac{1}{x} > 1$, we have $x^y + y^x > 1$
