Proving the exponential inequality: $x^y+y^x\gt1$

Question

How can the following inequality be proven? $$x^y+y^x\gt1$$ for $(x\gt0,y\lt1)$

Answer 1

Use this Bernoulli inequality $$(1+x)^a\le 1+ax,0<a<1,x>-1$$

It is clear we only prove $0<x,y<1$ then $$x^y=\dfrac{1}{\left(\dfrac{1}{x}\right)^y}=\dfrac{1}{\left(1+\dfrac{1-x}{x}\right)^y}>\dfrac{1}{1+\dfrac{1-x}{x}\cdot y}=\dfrac{x}{x+y-xy}>\dfrac{x}{x+y}$$ Similarly $$y^x>\dfrac{y}{x+y}$$ so $$x^y+y^x>\dfrac{x}{x+y}+\dfrac{y}{x+y}=1$$

Answer 2

I see that $x^y>\frac{x}{x+y}$ for $\{x,y\}\subset(0,1)$ by another way.

We need to prove that $f(x)>0$, where $f(x)=x^y+yx^{y-1}-1$.

Indeed, $f'(x)=yx^{y-1}+y(y-1)x^{y-2}=yx^{y-2}(x+y-1)$, which gives $x_{min}=1-y$.

Id est, $f(x)\geq f(1-y)=(1-y)^{y-1}(1-y+y)-1=(1-y)^{y-1}-1>0$

and we are done!

Answer 3

Hint: for $x\ge 1$ the proof is trivial. For $0 \lt x,y \lt 1$, the function is symmetric in $x$ and $y$, so it is same as finding minimum of $x^x$ in interval (0,1).