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

Explanation

Since, AM of 2nd powers gt 2nd power of AM
`:. ((a+(1)/(a))^(2)+(b+(1)/(b))^(2))/(2)gt((a+(1)/(a)+b+(1)/(b))/(2))^(2)`
`= (1)/(4)(a+b+a^(-1)+b^(-1))^(2)=(1)/(4)(1+a^(-1)+b^(-1))^(2) " "[:. a+b=1]`
`:.(a+(1)/(a))^(2)+(b+(1)/(b))^(2)gt(1)/(2)(1+a^(-1)+b^(-1))^(2)" " "......(i)"`
Again, `(a^(-1)+b^(-1))/(2)gt((a+b)/(2))^(-1)=((1)/(2))^(-1)=2`
or `(a^(-1)+b^(-1))/(2)gt2`
`implies a^(-1)+b^(-1)gt4`
`:. (1+ a^(-1)+b^(-1))gt5 " or " (1+ a^(-1)+b^(-1))^(2)gt25`
` implies (1)/(2) (1+ a^(-1)+b^(-1))^(2)gt(25)/(2) " " ".....(ii)"`
From Eqs. (i) and (ii), we get
` (a+(1)/(a))^(2)+(b+(1)/(b))^(2)gt(25)/(2)`.