Demonstrate that: 3^sqrt 5

We'll apply the 4th consequence of the Lagrange's
theorem.

In fact, we'll have to prove
that:

n^sqrt(n+2) < (n+2)^sqrt
n

We'll take natural logarithms both
sides:

ln n^sqrt(n+2) < ln (n+2)^sqrt
n

We've taken natural logarithms both sides, to have a base
of logarithms (the base is e = 2.7...) that is > 1 and the direction of the
inequality to hold.

We'll use the power property of
logarithms:

sqrt(n+2)*ln n < sqrt n*ln
(n+2)

We'll divide both sides by sqrt
n:

sqrt(n+2)*ln n/sqrt n < ln
(n+2)

We'll divide both sides by sqrt
(n+2):

ln n/sqrt n < ln (n+2)/sqrt
(n+2)

We'll take the function f(x) = ln x/sqrt x, where x
is in the interval [3 ; 5].

If we demonstrate that the
function is increasing over the interval [3 ; 5], the inequality is
true.

A function is increasing if and only if, for x =
3<x = 5, we'll get:

f(3) <
f(5)

Let's prove that the function is increasing. For this
to hapen, the derivative of the function has to be positive. We'll do the derivative
test:

f'(x) = (ln x/sqrt
x)'

We'll apply the quotient
rule:

(u/v)' =
(u'*v-u*v')/v^2

u = ln x, u' =
1/x

v = sqrt x , v' = 1/2sqrt
x

(ln x/sqrt x)' = (sqrtx/x - ln
x/2sqrtx)/x

(ln x/sqrt x)' = (2sqrtx -
sqrtx*lnx)/2x^2

It is obvious that (ln x/sqrt x)' >
0 over the interval [3 ; 5], so f(x) is increasing.

ln
3/sqrt 3 < ln 5/sqrt 5