<s>[INST] Prove the inequality $(1-a)^b>(1-b)^a$ if $1>b>a>0$ by using binomial expansion For this question Proving or disproving: If $0<a<b<1$, then $(1-a)^b>(1-b)^a$ it is already proved. But now I want to proof it using binomial expansion, please help to verify if the proof valid.
Given $1>b>a>0$ and let $x$ bigger or equal to 1 then
$\frac{x-a}{a}>\frac{x-b}{b}$
According to binomial expansion
$(1-a)^b=1-ab+\frac{b(b-1)}{2!}(-a)^2+\frac{b(b-1)(b-2)}{3!}(-a)^3+...$
and
$(1-b)^a=1-ab+\frac{a(a-1)}{2!}(-b)^2+\frac{a(a-1)(a-2)}{3!}(-b)^3+...$
I compare each nth term and multiply the nth term by $\frac{n!}{(ab)^n}$ to simplify the evaluation
Example for the 2nd term of $(1-a)^b$
$\frac{b(b-1)}{2!}(-a)^2.(\frac{2!}{(ab)^2})=\frac{-1(1-b)}{b}$
and 2nd term of $(1-b)^a$
$\frac{a(a-1)}{2!}(-b)^2.(\frac{2!}{(ab)^2})=\frac{-1(1-a)}{a}$
From above inequality
$\frac{1-a}{a}>\frac{1-b}{b}$ times negative one then $\frac{-1(1-b)}{b}>\frac{-1(1-a)}{a}$ which show the 2nd term of $(1-a)^b$ from binomial expansion bigger than $(1-b)^a$
Repeat the procedure and for 3rd term will get
$\frac{-1(1-b)(2-b)}{b^2}>\frac{-1(1-a)(2-a)}{a^2}$ since from above equality $\frac{1-a}{a}>\frac{1-b}{b}$ and $\frac{2-a}{a}>\frac{2-b}{b}$ it follow that $\frac{1-a}{a}.\frac{2-a}{a}>\frac{1-b}{b}.\frac{2-b}{b}$
This proof can be repeat for subsequent term which conclude the proof $(1-a)^b>(1-b)^a$ since all the nth term in $(1-a)^b>(1-b)^a$
 [/INST] You're on the right track. You have two base cases and now you need to prove your claim that it holds for the $n^{th}$ term using induction.
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