# Deduction of an inequality involving $\int_2^x\frac{dt}{\log t}$ and $\int_2^x\frac{dt}{\log^2 t}$

I try repeat the PROBLEMA 49, see page 149, here in spanish, to obtain an inequality, if it is possible, involving the logarithmic integrals $Li(x)=\int_2^x\frac{dt}{\log t}$ and $Li_2(x)=\int_2^x\frac{dt}{\log^2 t}$. The outline of cited problem is very easy: first use Schwartz inequality for some interesting functions, secondly define a function $F(x)$ such that $F'(x)\geq \text{a constant}$, in an interval $[y,x]$ that provides us combining it with previous computations as $$F(x)-F(y)\geq (\text{a constant})\cdot(x-y)$$ the inequality deduced by Rolle's theorem.

Question. Is it possible following the outline to deduce an inequality of this kind (like the inequality deduced in cited problem) involving $Li(x)$ and $Li_2(x)$? Thanks in advance.

My attempt was first by Schwartz inequality $$(Li(2+a))^2\leq a\cdot Li_2(2+a)$$ that holds for every $a>0$. After I define (you can take other definition) $$F(x):=\frac{Li\left(2+\frac{1}{x}\right)}{Li_2\left(2+\frac{1}{x}\right)},$$ with derivative (if there are no mistakes) $$F'(x)=-\frac{1}{x^2\log^2\left(1+\frac{1}{x}\right)Li\left(2+\frac{1}{x}\right)^2}+\frac{2Li_2\left(2+\frac{1}{x}\right)}{x^2\log\left(1+\frac{1}{x}\right)Li\left(2+\frac{1}{x}\right)^3},$$ thus combining with Schwartz inequality that is $$\frac{1}{x}Li_2\left(2+\frac{1}{x}\right)\cdot\frac{1}{Li\left(2+\frac{1}{x}\right)^2}\geq 1$$ one needs to prove or refute, if it is false, the following inequality

$$-\frac{1}{x^2\log^2\left(1+\frac{1}{x}\right)Li\left(2+\frac{1}{x}\right)^2}+\frac{2}{x\log\left(1+\frac{1}{x}\right)Li\left(2+\frac{1}{x}\right)}\geq \text{a constant}.$$ I don't know how to discuss the sign of this last inequality.

I don't know if I understood your question (if my interpretation is wrong tell me and I will delete the answer) but note that, integrating by parts, $$\int_{2}^{x}\frac{dt}{\log\left(t\right)}=\frac{x}{\log\left(x\right)}-\frac{2}{\log\left(2\right)}+\int_{2}^{x}\frac{dt}{\log^{2}\left(t\right)}$$ and so $$\textrm{Li}\left(x\right)-\textrm{Li}_{2}\left(x\right)=\frac{x}{\log\left(x\right)}-\frac{2}{\log\left(2\right)}$$ then if we define $$F\left(x\right)=\log\left(x\right)\left(\textrm{Li}\left(x\right)-\textrm{Li}_{2}\left(x\right)+\frac{2}{\log\left(2\right)}\right)$$ we have $$F\left(x\right)-F\left(y\right)=x-y.$$ Also note that $$F'\left(x\right)=\frac{\textrm{Li}\left(x\right)-\textrm{Li}_{2}\left(x\right)+\frac{2}{\log\left(2\right)}}{x}+1-\frac{1}{\log\left(x\right)}$$ and since $$\textrm{Li}\left(x\right)\sim\frac{x}{\log\left(x\right)},\,\textrm{Li}_{2}\left(x\right)\sim\frac{x}{\log^{2}\left(x\right)}$$ we have $$F'\left(x\right)\geq C$$ for some $C>0$.
Addendum: I think it's useful to underline that the calculations about the derivative are only for “exercise” since $F\left(x\right)=x$ and so $F'\left(x\right)=1$.