Question

If $a,b,c$ are in descending order of magnitude, show that
${\left(\frac{a+c}{a-c}\right)}^a<{\left(\frac{b+c}{b-c}\right)}^b$.

Answer 1

Given $a>b>c$

Let , $a=\frac{c}{y}$ and $b=\frac{c}{x}$

As ,$a>b$ ,

$\frac{c}{y}>\frac{c}{x}$

$\frac{1}{y}>\frac{1}{x}$

$x>y$

${\left(\frac{a+c}{a-c}\right)}^a=\left(\frac{\frac{c}{y}+c}{\frac{c}{y}-c}\right)={\left(\frac{1+y}{1-y}\right)}^{\frac{c}{y}}$

${\left(\frac{b+c}{b-c}\right)}^b=\left(\frac{\frac{c}{x}+c}{\frac{c}{x}-c}\right)={\left(\frac{1+x}{1-x}\right)}^{\frac{c}{x}}$

As $x>y$ then

${\left(\frac{1+x}{1-x}\right)}^{\frac{1}{x}}>{\left(\frac{1+y}{1-y}\right)}^{\frac{1}{y}}$

${\left(\frac{1+x}{1-x}\right)}^{\frac{c}{x}}>{\left(\frac{1+y}{1-y}\right)}^{\frac{c}{y}}$