Consider the system of equations `x_1+x_(2)^(2)+x_(3)^(3)+x_(4)^(4)+x_(5)^(5)=5` and `x_1+2x_2+3x_3+4x_4+5x_5=15` where `x_1,x_2,x_3,x_4,x_5` are positive real numbers. Then numbers of `(x_1,x_2,x_3,x_4,x_5)` is ___________.

Answer

1

Explanation

Using A.M. `ge` G.M., We get
`x_(2)^(2)+1 ge 2x_(2)`
`x_(3)^(3)+1+1ge 3x_3`
`x_(4)^(4)+1+1+1ge 4x_(4)`
`x_(5)^(5)+1+1+1+1ge 5x_(5)`
Adding these, we get
`x_1+x_(2)^(2)+x_(3)^(3)+x_(4)^(4)+x_(5)^(5) +10ge x_1+2x_2+3x_3+4x_4+5x_5`
`x_1+x_(2)^(2)+x_(3)^(3)+x_(4)^(4)+x_(5)^(5)=5`
The equality holds if `x_1=x_2=x_3=x_4=x_5=1` .
So, there is only one posibility for `(x_1x_2,x_3,x_4,x_5)`.