I think it gives a better sense of the geometry to ask whether, with non-negative $x,y$ such that $$ \frac{1}{2} \leq x + y \leq 1, $$ can we prove that $$ x^{2 y} + y^{2 x} \leq 1 ?$$ I'm not entirely certain where the second level curve component, through $\left( \frac{1}{4} , \frac{1}{4}\right),$ meets the axes. My programmable calculator seems to think that, if this arc does have $\left( \frac{1}{2} , 0 \right)$ as a limit point, the arc is tangent to the $x$-axis.

I see, this was pointed out in a comment on March 17 by Yaakov Baruch, one needs to click on the "show 6 more comments." I think I will leave this here anyway.

I think it gives a better sense of the geometry of the problem to ask whether, with non-negative $x,y$ such that $$ \frac{1}{2} \leq x + y \leq 1, $$ we can prove that $$ x^{2 y} + y^{2 x} \leq 1 ?$$ I'm not entirely certain where the second level curve component, through $\left( \frac{1}{4} , \frac{1}{4}\right),$ meets the axes. My programmable calculator seems to think that, if this arc does have $\left( \frac{1}{2} , 0 \right)$ as a limit point, the arc is tangent to the $x$-axis.

I see, this was pointed out in a comment on March 17 by Yaakov Baruch, one needs to click on the "show 6 more comments." I think I will leave this here anyway.

I think it gives a better sense of the geometry to ask whether, with non-negative $x,y$ such that $$ \frac{1}{2} \leq x + y \leq 1, $$ can we prove that $$ x^{2 y} + y^{2 x} \leq 1 ?$$ I'm not entirely certain where the second level curve component, through $\left( \frac{1}{4} , \frac{1}{4}\right),$ meets the axes. My programmable calculator seems to think that, if this arc does have $\left( \frac{1}{2} , 0 \right)$ as a limit point, the arc is tangent to the $x$-axis.

I think it gives a better sense of the geometry to ask whether, with non-negative $x,y$ such that $$ \frac{1}{2} \leq x + y \leq 1, $$ can we prove that $$ x^{2 y} + y^{2 x} \leq 1 ?$$ I'm not entirely certain where the second level curve component, through $\left( \frac{1}{4} , \frac{1}{4}\right),$ meets the axes. My programmable calculator seems to think that, if this arc does have $\left( \frac{1}{2} , 0 \right)$ as a limit point, the arc is tangent to the $x$-axis.

I see, this was pointed out in a comment on March 17 by Yaakov Baruch, one needs to click on the "show 6 more comments." I think I will leave this here anyway.