If `a , b >0a n d0

Explanation

Let `f(x) =(1+x)^((p-1)-x^(p)),xgt0`
`therefore f(x)=p(1+x)^(p-1-px^(p-1))`
`=p[(1+x)^(p-1)-x^(p-1)]`
Now `1+xgt`
`(1+x)^(1-p)gtx^(1-p)`
`(1)/(1+x)^(p-1)gt(1)/(x^(p)-1)`
`(1+x^(p-1)ltx^(p-1))`
`(1+x)^(p-1)-x^(p-1)lt0`
From (1) and (2) we get f(x) `lt` 0
So f(x) is a decreasing function f
Now f(0) =0
`xgt0`
`f(x)ltf(0)`
`(1+x)^(p-1)-x^(p)lt0`
`(1+x)^(p)lt1+x^(p)`
Putting x=`(a)/(b)`. ,we get `(a+b)^(p)lta^(p)+b^(p)`