Question on mean value theorem please help

Prove that
1/9 is greater then or equal to
square root(66) - 8 is less then or equal to 1/8
(without computing square root(66))?

Apply the Mean Value Theorem to f(x) = √x for x in [64, 66].

f(66) - f(64)/(66 - 64) = f '(c) for some c in (64, 66).
==> (√66 - 8)/2 = 1/(2√c) for some c in (64, 66).

However, since g(x) := 1/(2√x) is a decreasing function for x > 0, we have
0 < 1/(2√c) < 1/(2√64) for all c > 64.

Hence, 0 < (√66 - 8)/2 < 1/(2√64)
==> 0 < √66 - 8 < 1/8.

I hope this helps!