# A question about the second Chebyshev function $\psi(x) = \sum_{m=1}^{\infty}\vartheta(\sqrt[m]{x})$

Using a simple java application, I have noticed that for $x > 25$:

$$\psi\left(\frac{x}{5}\right) \ge \psi\left(\frac{x}{3}\right) - \psi\left(\frac{x}{4}\right)$$

where:

$$\psi\left(x\right) = \sum_{m=1}^{\infty}\vartheta\left(\sqrt[m]{x}\right)$$

and

$$\vartheta\left(x\right) = \sum_{p \le x} \log p$$

As I understand it, based on the Prime Number Theorem, this inequality should be true.

Is this inequality elementary and straight forward to prove? Or is this more difficult to prove than it appears?

Thanks very much,

-Larry

## 1 Answer

The inequality is simpler than PNT but also I would not consider it as straightforward; what seems to be needed are results slightly weaker than those of Chebyshev.

Suppose we can bound $$c_1 y < \psi(y) < c_2 y$$. Then for your inequality it would suffice to have $$c_1/5 \ge c_2/3 - c_1/4$$
which translates to $$c_1 \frac{27}{20} \ge c_2$$ Now Chebyshev established such bound with $$c_1$$ (about $$0.92$$) and $$c_2 = c_1 \frac{6}{5}$$, except for some small explicit logarithmic terms, for $$y \ge 30$$.

Thus starting from quite small $$x$$ your inequality will already follow from Chebyshev's work.