# Use mean value theorem to show that for 0<x<y, sqrty-sqrtx<y-x/2sqrtx

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.

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