Question

If $a,b>0$ and $0<p<1,$ then which of the following option is CORRECT?

• A. $(a+b{)}^p>a^p+b^p$
• B. $(a+b{)}^p<a^p+b^p$
• C. $(a+b{)}^p\ge a^p+b^p$
• D. $(a+b{)}^p\le a^p+b^p$

Answer 1

The correct option is B $(a+b{)}^p<a^p+b^p$

$\text{Let,}f\left(x\right)=(1+x{)}^p-1-x^p,x>0$
$\Rightarrow f^{\prime }\left(x\right)=p(1+x{)}^{p-1}-p{x}^{p-1}\cdots (1)$
Now,
$1+x>x$
$\Rightarrow (1+x{)}^{1-p}>x^{1-p}$
$\Rightarrow \frac{1}{(1+x{)}^{p-1}}>\frac{1}{x^{p-1}}$
$\Rightarrow (1+x{)}^{p-1}<x^{p-1}$
$\Rightarrow (1+x{)}^{p-1}-x^{p-1}<0\cdots (2)$
From $\left(1\right)$ and $\left(2\right),$ we get
$f^{\prime }\left(x\right)<0$
Therefore, $f\left(x\right)$ is a decreasing function
Now, $f\left(0\right)=0$
$\because x>0\Rightarrow f\left(x\right)<f\left(0\right)$
$\Rightarrow (1+x{)}^p-1-x^p<0$
$\Rightarrow (1+x{)}^p<1+x^p$
Put $x=\frac{a}{b},$
Hence, $(a+b{)}^p<a^p+b^p$