Question

If a , b are = ive , then from any integer n prove that
$(a+b{)}^n<2^n(a^n+b^n)$

Answer 1

(a + b ) < 2 (a + b )
$\therefore$ p(1) holds goods
Assume p (m) i. e.
$(a+b{)}^m<2^m(a^m+b^m)$
We have to prove p( m + 1)
i. e. $(a+b{)}^{m+1}<2^{m+1}(a^{m+1}+b^{m+1})$
now $(a+b{)}^{m+1}=(a+b{)}^m+(a+b)$
$<2^m(a^m+b^m)(a+b)$
$=[a^{m+1}+b^{m+1}+a^{m}b{b}^{m}a]$
$<2^m[a^{m+1}+b^{m+1}{a}^{m+1}+b^{m+1}]$
as shown below in
$=2^m[a^{m+1}+b^{m+1}]$
now $a^{m}b+b^{m}a<a^{m+1}+b^{m+1}$
or $a^m(b-a)+b^m(a-b)<0$
or $(a^m-b^m)(b-a)>0$
or $(a^m-b^m)(a-b)>0$
above is true because when a and b are both +ive then either a < b or a > b
When a < b $a^m-b^m$ < o , a - b < 0
$\therefore$ (b) holds
When a > b i. e.
$a^m-b^m$ < o , a - b < 0
$\therefore$ (b) holds
$\therefore$ p ( m + 1) holds goods hence universally true