Proving the exponential inequality: $x^y+y^x\gt1$
Solution 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 $$x^y+y^x>\dfrac{x}{x+y}+\dfrac{y}{x+y}=1$$