Question

If $a>b$, show that $a^b{b}^b>a^b{b}^a$, and $\log \frac{b}{a}<\log \frac{1+b}{1+a}$.

Answer 1

Given $a>b$

$b-a<0$

Multiplying with $\log b$ both sides we get

$(b-a)\log b<0b\log b-a\log b<0\log b^b<\log b^a$

Adding $\log a^b$ both sides we get

$\log a^b+\log b^b<\log a^b+\log b^a\log a^b{b}^b<\log a^b{b}^a{a}^b{b}^b<a^b{b}^a$

Given $a>b$

Adding $ab$ both sides we get

$ab+a>ab+ba(1+b)>b(1+a)\frac{(b+1)}{(1+a)}>\frac{b}{a}$

Taking $\log$ both sides we get

$\log \frac{b}{a}<\log \frac{1+b}{1+a}$