Prove that log 5 (6)+log 6 (7)+log 7 (8)+log 8 (5)>4I've put the base the first and the number into bracket

log 5 (6) + log 6 (7) + log 7 (8) + log 8 (5) >
4

We know that log a (b) = log c (b) / log c
(a)

Let c be 10 ==> log a (b) = log b/ log
a

Then, let us rewrite:

log 6/
log 5 + log 7/ log 6 + log 8 / log 5

Since log x is an
increasing function, then f(x+1) > f(x) , then f(x+1)/f(x) >
1

==> log 6/ log 5 >
1.......(1)

==> log 7/ log 6 >
1.......(2)

==> log 8/ log 7 >
1.......(3)

==> log 5 / log 8 >
1......(4)

Now add (1) and (2) and (3) and
(4)

==> log 6/log5 + log 7/log6 + log 8/log 7 +
log5/ log 8 > 3

==> log 5 (6) + log 6 (7) +
log 7 (8) + log 8 (5) > 4