If` a^(2) +b^(2) prop ab`, then prove that (a+b) prop (a-b).

Explanation

`(a^(2)+b^(2)) prop `ab.
`rArr (a^(2) +b^(2) )=k.ab [when k ne 0 "variation constant"]`
` therefore a^(2)+b^(2) =k.ab........(1)`
`rArr(a+b)^ (2)-2ab=k.ab`
`(a+b)^ (2)=kab+2ab`
`rArr (a+b)^(2) =(k+2)ab …….(2)`
Again , `a^(2) +b^(2) =k.ab`
`rArr a^(2) +b^(2) -2ab=k.ab-2ab`
`rArr (a-b)^(2) =(k-2) ab .......(3)`
Now , dividing (2) by (3) we get , `(a+b)^(2)/(a-b)^(2) =((k+2)ab)/((k-2)ab)`
` or (a+b)^(2)/(a-b)^(2)=(k+2)/(k-2)=m ("let")ne0 [because(k+2)/(k-2)ne0]`
or ,` (a+b)^(2) =m (a-b)^(2) `
or, a+b =`sqrt(m) (a-b)`
`therefore` (a+b)`prop (a-b) [because sqrt(m) ne0= ` variation constant]
Hence (a+b) `prop` (a-b) . (Proved )