I want to prove the following inequality:
$$ \int _2 ^x \frac{dt}{\log t} \leq \frac{\sqrt{x}}{\log 2} + \frac{x - \sqrt{x}}{\log \sqrt{x}}$$
I was able to prove, for $t>-1,$ that $$ \frac{t}{1+t} < \log (1+t) < t $$
but I am unsure on what to do next as this expression does not involve square roots. We also know that $\log x < \sqrt{x}.$ Any suggestions on how to proceed/conclude? (I am trying to prove this inequality for use in Number Theory.)
I assume that you mean $ x\ge 2$
This integral is $Li(x)+C$
By substitution that $T=\log t$, and integration by part, becoming
$$ x^{-1}e^x-\int e^x d(x^{-1}) $$
Substituting, we get $$\frac{x}{\log x} - \frac{e^2}{\log 2} -C(x)$$ $C(x)$ is the second integral that is in positive value, as for $ x\ge 2$, $x^{-2}e^x \ge 0$ .
Also, $\frac{e^2}{\log 2} \gt 0$
As we know for $x \ge 4 $,
$\frac{\sqrt{x}}{\log 2}-\frac{\sqrt{x}}{\log \sqrt{x}} \ge 0$
So the inequality becomes, $$\frac{x}{\log x} - C_1 \le \frac{x}{\log x} + C_2$$ , where $C_1,C_2$ are positive constants.
For $ 2 \lt x \lt 4 $, in
$$\frac{\sqrt{x}}{\log 2}-\frac{\sqrt{x}}{\log \sqrt{x}} $$ since $\frac{1}{\log \sqrt x} =\frac{2}{\log x} \le \frac{2}{\log 2 }$,
as such, the expression becomes $ \frac {-\sqrt x}{\log 2} \ge \frac{-2}{\log 2}$, which is again $ \ge \frac{-e^2}{\log 2}$,
so the inequality still hold.