You need to rememeber what the mean value theorem tells such that:
`f'(c) = (f(b) - f(a))/(b - a)`
`f:[a,b]-gtR, c in [a,b]`
f continuous over [a,b]
Hence, you need to consider the function `f(z)=sqrt z` , f continuous over (x,y) such that:
`f'(z) = (sqrty - sqrtx)/(y-x)`
`(sqrt z)' = (sqrty - sqrtx)/(y-x) =gt 1/(2sqrtz) = (sqrty - sqrtx)/(y-x)`
`(y - x)/(2sqrt z) =(sqrty - sqrtx)`
You need to remember that `z in (x,y) =gt xltzlty`
`sqrtx lt sqrt z lt sqrt y`
`2sqrt x lt 2sqrt z lt 2sqrt y`
`1/(2sqrt x)gt 1/(2sqrt z)gt 1/(2sqrt y)`
Since `y gt x =gt y - x gt 0` , hence if you multiply by y - x inequality above yields:
`(y-x)/(2sqrt x) gt (y-x)/(2sqrt z) gt (y-x)/(2sqrt y)`
Notice that `(y - x)/(2sqrt z) = (sqrty - sqrtx)` by Mean Value Theorem such that:
`(y-x)/(2sqrt x) gt (sqrty - sqrtx)`
Hence, by Mean Value Theorem yields that `(sqrty - sqrtx) lt (y-x)/(2sqrt x)` under given conditions.