Question

Let $a>b>0$ and $I\left(n\right)=a^{1/n}–b^{1/n},$ $J\left(n\right)=(a–b{)}^{1/n}$ for all $n\ge 2$. then

• A. $I\left(n\right)<J\left(n\right)$
• B. $I\left(n\right)>J\left(n\right)$
• C. $I\left(n\right)=J\left(n\right)$
• D. $I\left(n\right)+J\left(n\right)=0$

Answer 1

The correct option is A $I\left(n\right)<J\left(n\right)$

For $x>0,y>0$
$(x+y{)}^n>x^n+y^n$ when $n>1$
$\Rightarrow (x+y{)}^n<x^n+y^n$ when $0<n<1$
$\Rightarrow x^{1/n}+y^{1/n}>(x+y{)}^{1/n}$ when $n>1$

Taking $x=b$ and $y=a-b$
$\Rightarrow b^{1/n}+(a-b{)}^{1/n}>(b+(a-b){)}^{1/n}$
$\Rightarrow b^{1/n}+(a-b{)}^{1/n}>a^{1/n}$
$\Rightarrow (a-b{)}^{1/n}>a^{1/n}-b^{1/n}$
$\Rightarrow J\left(n\right)>I\left(n\right)$