# 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!

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 $$f(3) = e\log 2 - \log 3>0.$$

• Awesome! Love it. Thanks! – hello_123 Aug 9 at 0:20
• @hello_123: my pleasure! – user798202 Aug 9 at 0:20

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 $$x \geq 3$$.

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.