Question

If $x_1,x_2,x_3,x_4$ are four positive real numbers such that $x_1+\frac{1}{x_2}=4,x_2+\frac{1}{x_3}=1,x_3+\frac{1}{x_4}=4$ and $x_4+\frac{1}{x_1}=1,$ then

Answer 1

$Usingx+\frac{1}{y}\ge 2\sqrt{\frac{x}{y}},weget4=x_1+\frac{1}{x_2}\ge 2\sqrt{\frac{x_1}{x_2}}1=x_2+\frac{1}{x_3}\ge 2\sqrt{\frac{x_2}{x_3}}4=x_3+\frac{1}{x_4}\ge 2\sqrt{\frac{x_3}{x_4}}1=x_4+\frac{1}{x_1}\ge 2\sqrt{\frac{x_4}{x_1}}\Rightarrow 16=(x_1+\frac{1}{x_2})(x_2+\frac{1}{x_3})(x_3+\frac{1}{x_4})(x_4+\frac{1}{x_1})\ge 2^4$
Hence, each of the above inequality must be an equality.
Therefore, $x_1=\frac{1}{x_2},x_2=\frac{1}{x_3},x_3=\frac{1}{x_4}and{x}_4=\frac{1}{x_1}.\therefore x_1=2,x_2=\frac{1}{2},x_3=2,x_4=\frac{1}{2}$