Question

Let $a,b,c\in R^{+}$ such that $a+\frac{1}{b}=3,b+\frac{1}{c}=4,c+\frac{1}{a}=\frac{9}{11}$ then $abc=$

Answer 1

We have $a+\frac{1}{b}=3,b+\frac{1}{c}=4,c+\frac{1}{a}=\frac{9}{11}$

Adding all 3 equations

$a+b+c+(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=\frac{86}{11}$

Multiplying all 3 equations

$(a+\frac{1}{b})(b+\frac{1}{c}).(c+\frac{1}{a})=\frac{108}{11}$

On solving

$abc+(a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c})+\frac{108}{11}$

$\Rightarrow$ $abc+\frac{86}{11}=\frac{108}{11}-\frac{1}{abc}$

$\Rightarrow$ $abc+\frac{1}{abc}=\frac{22}{11}$

$\Rightarrow$ $abc+\frac{1}{abc}=2$

let $abc=x$

$\Rightarrow$ $x+\frac{1}{x}=2$

$\Rightarrow$ $x^2-2x+1=0$

$\Rightarrow$ ${(x-1)}^2=0$

$\Rightarrow$ $x=1$

$abc=1$