Show that [1+ln(x^x)+lnx]/(x+1)

We notice that the function f(x) = ln x is continuously and it
could be differentiated, we'll apply Lagrange's rule, over a closed interval [k ;
k+1].

According to Lagrange's rule, we'll
have:

f(k+1) - f(k) = f'(c)(k+1 - k)

c
belongs to the interval [k ; k+1].

f'(x) = 1/x => f'(c) =
1/c

We'll eliminate like terms inside brackets and we'll
have:

f(k+1) - f(k) = f'(c), where f'(c) =
1/c

f(k+1) - f(k) = 1/c

We'll re-write
the given inequality, over the closed interval [k ;
k+1].

[1+ln(k^k)+lnk]/(k+1)<ln(k+1)<[1+ln(k^k)]/k

We'll
apply the power rule of logarithms:

ln (k^k) = k*ln
k

[1+k*ln k+lnk]/(k+1)<ln(k+1)<[1+k*ln
k]/k

1/(k+1) + (lnk)*(k+1)/(k+1)<ln(k+1)< 1/k + k*(ln
k)/k

We'll simplify and we'll
get:

1/(k+1) + (lnk)<ln(k+1)< 1/k + (ln
k)

But ln k = f(k) and ln (k+1) =
f(k+1)

1/(k+1) + f(k) < f(k+1) < 1/k +
f(k)

We'll subtract f(k):

1/(k+1)
< f(k+1) - f(k) < 1/k

Since c belongs to [k ; k+1],
we'll write:

k < c <
k+1

1/k > 1/c >
1/(k+1)

But f(k+1) - f(k) = 1/c

1/k
> f(k+1) - f(k) > 1/(k+1) q.e.d.

According to
Lagrange's rule 1/k > f(k+1) - f(k) >
1/(k+1).

Based on the Lagrange's rule, the inequality
[1+ln(x^x)+lnx]/(x+1)<ln(x+1)<[1+ln(x^x)]/x is
verified.