Does the Prime Number Theorem imply for large enough $x$ that $(a+1)\pi(ax) \ge a\pi((a+1)x)$

Question

Give an integer $a \ge 1$, does the Prime Number Theorem imply for large enough $x$ that:

$$(a+1)\pi(ax) \ge a\pi((a+1)x)$$

where $\pi(x)$ is the prime counting function.

Using the formula of Pierre Dusart found here for $x > 598$:

$$\left(\frac{x}{\log x}\right)\left(1 + \frac{0.992}{\log x}\right) \le \pi(x) \le \left(\frac{x}{\log x}\right)\left(1 + \frac{1.2762}{\log x}\right)$$

this is true for $a=1$.

Given a large enough $x$, could it possibly be true for all $a$?

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Edit: Changed [] to () to avoid any confusion based on a question received in the comments.

Answer 1

Just use that $$ \pi (x) = \frac{x}{\log x} + O \left( \frac{x}{\log^3 x} \right) $$ (we can have more logs in the denominator of the error if we like). We thus have that $$ (a+1) \pi (ax) - a \pi ((a+1) x) $$ is roughly $$ \frac{ax}{\log^2 (ax)} $$ with error of order, at most, say, $$ O \left( \frac{a^2 x}{\log^3 (ax)} \right). $$ Taking $x$ suitably large relative to $a$ thus guarantees that the desired inequality holds.

Answer 2

We know that $ \frac{x}{\ln x}(1+\frac{1}{\ln x} +\frac{2}{\ln^2 x})\leq \pi(x) \leq \frac{x}{\ln x}(1+\frac{1}{\ln x}+ \frac{2.334}{\ln^2 x})$ for all $x\geq 3*10^{11}$

So for all $a\geq 1$ and $a \in \mathbb{R}$ we have that $ (a+1) \pi(a x)-a \pi((a+1)x) > (a+1) \frac{a x}{\ln a x} (1+\frac{1}{\ln ax }+\frac{2}{\ln^2a x})-a \frac{(a+1)x}{\ln(a+1)x}(1+\frac{1}{\ln(a+1)x}+\frac{2.334}{\ln^2 (a+1)x}) >0$ dividing by $a(a+1)$ to get

$ \frac{x}{\ln a x}(1+\frac{1}{\ln ax} +\frac{2}{\ln^2 ax})- \frac{x}{\ln(a+1)x}(1+\frac{1}{\ln(a+1)x}+\frac{2.334}{\ln^2(a+1)x})> (\frac{1}{\ln a x}-\frac{1}{\ln(a+1)x} )x (1+\frac{1}{\ln ax}+\frac{2}{\ln^2 ax })-\frac{0.334x}{\ln^3 (a+1)x} >\frac{\ln(1+1/a)}{\ln^2 (a+1)x} x-\frac{0.334 x}{\ln^3 (a+1)x}$

multiplying by $ \ln^2(a+1)x$ to get $ \ln(1+\frac{1}{a}) x > \frac{0.334x}{\ln a x} => \ln(1+1/a) > \frac{0.334}{\ln a x} => \frac{1}{a+0.5} > \frac{0.334}{\ln ax} $

giving $ \ln a x > 0.334(a+0.5) $ exponentiation both sides gives $ a x > e^{0.334 a+ 0.167} => x > \frac{1.2 e^{0.334a} }{a}$

So for every real number $a \geq 1$ the inequality is valid for all $ x \geq max(3*10^{11},\frac{1.2 e^{0.334a} }{a})$

Answer 3

You could also use the Inequality: $$\pi (x) + \pi (y) \ge \pi (x+y)$$

Using this, we can simply derive that: $$ (a+1) \pi(ax) \ge a \pi ((a+1)x))$$ $ (a+1) \pi(ax) = a \pi(ax) + \pi(ax)$

$$ a \pi(ax) + \pi(ax) \ge a \pi ((a+1)x)$$ $$ \pi (ax) + \frac {\pi (ax)}{a} \ge {\pi((a+1)x)}$$

Now since there can never be more primes than integers: $ \frac {\pi (ax)}{a} \le \pi(x) $

Using this, we can conclude:

$$ \pi (ax) + \pi(x) \ge \pi((a+1)x) = \pi (ax + x)$$

and now setting $y=ax$

$$ \pi(y) + \pi(x) \ge \pi(y+x)$$

Which proves the theorem.