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If a + b - 1, a gt 0,b gt 0, prove that ...

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

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Observe that `(x - y)^(2) ge 0 `
`rArr x^(2) + y^(2) - 2xy ge0 `
`rArr x^(2) + y^(2) ge 2xy`
`rArr (x^(2) + y^(2)) + (x^(2) + y^(2)) ge x^(2) + y^(2) + 2xy" "` [Adding `x^(2) + y^(2)` to each side ]
`rArr 2(x^(2) + y^(2)) ge (x + y)^(2)`
`rArr (x^(2) +y^(2))/(2) ge (x +y)^(2)/(4)`
`rArr(x^(2) +y^(2))/(2) ge ((x +y)/(2))^(2)`
That is , mean of squares is greater than or equal to square of mean.
Now applying this to the given sum
`((a+(1)/(a))^(2) + (b +(1)/(b))^(2))/(2) ge((a+(1)/(a) +b(1)/(b))/(2))^(2) ge ((a +b +(1)/(a) +(1)/(b))/(2))^(2)` ...(i)
a + b = 1 when plugged in
`(a + b ) ((1)/(a) +(1)/(b)) ge 4 ` given `(1)/(a) +(1)/(b) ge 4 `
From (i) ,
`((a + (1)/(a))^(2) +(b +(1)/(b))^(2) )/(2)ge ((1 +4)/(2))^(2)`
i.e., `(a+(1)/(a))^(2)+(b+(1)/(b))^(2) ge (25)/(2)`.
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