If a + b =1, a `gt` 0,b `gt` 0, prove that `(a + (1)/(a))^(2) + (b + (1)/(b))^(2) ge (25)/(2)`

Explanation

We know that A.M. of mth power `gt` mth power of A.M.
` therefore (((a+1)/(a))^2+(b+(1)/(b))^2)/(2)gt [((a+(1)/(b))^2+(b+(1)/(b)))/(2)]^2," here "m=2`
or ` (a+(1)/(a))^2+(b+(1)/(b))^2gt (1)/(2)[(a+b)+((1)/(a)+(1)/(b))]^2`
Also, `(a^-1+b^-1)/(2)gt ((a+b)/(2))^2`
or ` (1)/(2)((1)/(a)+(1)/(b))gt (2)/(a+b)`
or ` (1)/(a)+(1)/(b)gt (4)/(a+b)`
or ` (1)/(a)+(1)/(b)gt4`.
Hence, from (1)
From (i), ` (a+(1)/(a))^2+(b+(1)/(c))^2gt (1)/(2)(1+4)^2=(25)/(2)`