Proving that for all $x\geq 3$, $\log \log (x) \leq \log (\log(x-1)) + 1$? How do I go about proving that for all $x\geq 3$, $\log \log (x) \leq \log (\log(x-1)) + 1$?
When I differentiate to see if the lhs stays ahead, i lose the constant on the lhs and so i dont get anything meaningful. I also tried using some known inequalities like Jensen for concave functions but a naive application gives out an inequality in the other direction which is quite useless for this problem.
Any help is appreciated, thanks!
 A: 
How do I go about proving that for all $x\geq 3$, $\log \log (x) \leq \log (\log(x-1)) + 1$?

Assume in this answer that $\log$ means the natural logarithm with base $e$.
Since $\log A-\log B = \log\frac{A}{B}$, your inequality is equivalent to
$$
\log \frac{\log(x-1)}{\log (x)}=\log (\log(x-1))-\log \log (x)\ge -1=\log\frac{1}{e}\;,
$$
which is, by monotonicity of $\log$:
$$
\frac{\log(x-1)}{\log (x)}\ge \frac{1}{e}\;.
$$
So you want to show that for all $x\ge 3$:
$$
f(x) = e\log(x-1)-\log(x)\geq 0\;.
$$
Now, for all $x\ge 3$:
$$
f'(x) = \frac{e}{x-1}-\frac{1}{x} = \frac{(e-1)x+1}{x(x-1)}\;>0
$$
But
$$
f(3) = e\log 2 - \log 3>0.
$$
A: An easier approach is possible :
$\log(\log(x)) \leq \log(\log(x-1))+1 \implies \log(x) \leq e \cdot log(x-1) \implies x \leq (x-1)^e $
From here you can use derivatives of $x$ and $(x-1)^e$ to prove that the inequality is true. Indeed, the inequality is verified in 3 and the derivative of rhs is always greater when $x \geq 3$.
A: Not sure how much this solution will help you; it's a relatively simple, elementary method accessible to your usual Calculus I student, as opposed to appealing to more "advanced" ideas like Jensen's inequality. Still, hopefully it proves useful.

Raise both sides to the $e$ twice. After the  first,
$$\log(x) \stackrel{(?)}\le e\log(x-1)$$
Do it again, then
$$x \stackrel{(?)}\le  e^{e \log(x-1)} = (e^{\log(x-1)})^e = (x-1)^e$$
Thus, $x \le (x-1)^e$ is an equivalent inequality to our given one. Or, even more useful, $f(x) := x - (x-1)^e \le 0$ is an equivalent one.

Notice that $f'(x) = 1 - e(x-1)^{e-1}$. If we set $f'(x) = 0$, then we see that
$$x = 1 + \left( \frac{1}{e} \right)^{1/(e-1)} \approx 1.56$$
which is the only such zero for $f$: $f(x) > 0$ for $x$ to the left, and $f(x) < 0$ for $x$ to the right.
So this essentially means $f$ has a roughly "parabolic down" shape. We want to ensure $f(x) \le 0$ whenever $x \ge 3$. We can, in fact, do even better. When is $f(x) = 0$? Checking the graph suggests it's about $2.3$; checking the easier $x=2.5$, for instance, we see $f(x) < 0$ there ($f(2.5) \approx -0.51$). And of course you can check $f(2)$ to see $f(2) = 1 > 0$, which ensures that $f(x) = 0$ for some $x \in (2,2.5)$ by the intermediate value theorem.
Since $f'(x) < 0$ for $x \gtrsim 1.56$, we're ensured there will be no zeroes $x \gtrsim 1.56$ as well. (After all, $f$ is continuous and differentiable on its domain, and its derivative has only the one real root. Being able to become positive again and violate the inequality would require that there be a "turning point" where $f'(x)=0$, or that $f$ suddenly "jumps" to above the $x$-axis.)
Thus, we know $f(x) := x - (x-1)^e \le 0$ whenever $x \ge 2.5$. We can return to our original inequality by reversing our steps: bring the $(x-1)^e$ to the other side, then take the logarithm of each side twice.
