a two-variable cyclic power inequality $x^y+y^x>1$ intractable by standard calculus techniques

Question

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.

Answer 1

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 $x^y + y^x > 1$

Answer 2

let $x,y\in (0,1)$ and let $y^{\frac{1}{1-y}}\geq x^{\frac{1}{1-x}}$ then we have $\frac{x^{y-1}}{y^{x-1}}\geq 1$ Now let $S=(x^y+y^x)^{1/x}$ since $\frac{1}{x}>1$ we get $(1+x)^{r}\geq 1+rx$ if $x>-1,r\geq1$ and we have $S\geq y(1+\frac{x^{y-1}}{y^x})=y+\frac{x^{y-1}}{y^{x-1}}\geq y+1$ therefore we have $S\geq 1+y>1$ and since $S>1$ we get $S^x>1$

Answer 3

Why it's intractable by standard calculus?

If $x\geq1$ or $y\geq1$ so the inequality is obviously true.

Thus, we can assume $\{x,y\}\subset(0,1)$.

We'll prove that $x^y\geq\frac{x}{x+y}$.

Indeed, $x^y\geq\frac{x}{x+y}\Leftrightarrow f(x)\geq1$, where $f(x)=x^y+yx^{y-1}$.

$f'(x)=yx^{y-1}+y(y-1)x^{y-2}=yx^{y-2}(x+y-1)$.

Id est, $x_{min}=1-y$ and $f(x)\geq f\left(x_{min}\right)=(1-y)^{y-1}>1$.

Hence, also $y^x\geq\frac{y}{x+y}$ and $x^y+y^x\geq1$. Done!