Exponential teaser [closed]

1. Find the minimum value for $x^x$ for $x$ a positive real number

2. If $x$ and $y$ are positive real numbers, show that $y^x + x^y > 1$

Solution 1:

1. $\frac{d}{dx}{x^x}=x^x(\ln(x)+1)$. Thus $x^x$ is decreasing for $x<\frac{1}{e}$, increasing for $x>\frac{1}{e}$, and attains its minimum of $e^{-\frac{1}{e}}$ at $x=\frac{1}{e}$.

2. If $x \geq 1$, then $x^y+y^x>x^y \geq 1$.
   Similarly if $y\geq 1$, then $x^y+y^x>y^x \geq 1$.
   Finally if $0<x, y<1$, let $x=\frac{1}{1+a}, y=\frac{1}{1+b}, 0<a, b$, so that the inequality becomes $\frac{1}{(1+a)^y}+\frac{1}{(1+b)^x}>1$. By Bernoulli's inequality, since $0<x, y<1$ and $0<a, b$, we have $(1+a)^y \leq 1+ya$ and $(1+b)^x \leq 1+xb$. Thus $\frac{1}{(1+a)^y}+\frac{1}{(1+b)^x} \geq \frac{1}{1+ya}+\frac{1}{1+xb}=\frac{2+a+b}{1+a+b}>1$