Write this as

ln(x+1)/ln(x) <= 1 + 1/(x ln x).

Since x > 1, we have ln x > 0. So we can multiply both sides by ln x and obtain an equivalent inequality:

ln (x + 1) <= ln x + 1/x, or

ln(x + 1) - ln x <= 1/x.

Here there are a couple of options.

Method 1. Rewrite the equation as ln(1 + 1/x) <= 1/x. Then it results from the inequality ln(1 + t) <= t.

This last inequality can be proved by noting that h(t) = t - ln(1 + t), defined for t > -1, satisfies h(0) = 0 and h'(t) = t/(1 + t). Since 1 + t > 0, we have h'(t) < 0 for t < 0 and h'(t) > 0 for t > 0. Thus h is decreasing on (-1,0\] and increasing on \[0,+inf), which proves h(t) >= h(0) = 0 for all t.

Method 2. The left-hand side is equal to the integral of dt/t from x to x + 1, while the right-hand side is the integral of dt/x from x to x + 1. But 1/t <= 1/x for t in \[x,x+1\].