Vtyjkyuk

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

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

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

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

$$left(fracxlog xright)left(1 + frac0.992log xright) le pi(x) le left(fracxlog xright)left(1 + frac1.2762log xright)$$

this is true for $a=1$.

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

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

Just use that
$$
pi (x) = fracxlog x + O left( fracxlog^3 x right)
$$
(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
$$
fracaxlog^2 (ax)
$$
with error of order, at most, say,
$$
O left( fraca^2 xlog^3 (ax) right).
$$
Taking $x$ suitably large relative to $a$ thus guarantees that the desired inequality holds.

We know that $ fracxln x(1+frac1ln x +frac2ln^2 x)leq pi(x) leq fracxln x(1+frac1ln x+ frac2.334ln^2 x)$ for all $xgeq 3*10^11$

So for all $ageq 1$ and $a in mathbbR$ we have that $ (a+1) pi(a x)-a pi((a+1)x) > (a+1) fraca xln a x (1+frac1ln ax +frac2ln^2a x)-a frac(a+1)xln(a+1)x(1+frac1ln(a+1)x+frac2.334ln^2 (a+1)x) >0$ dividing by $a(a+1)$ to get

$ fracxln a x(1+frac1ln ax +frac2ln^2 ax)- fracxln(a+1)x(1+frac1ln(a+1)x+frac2.334ln^2(a+1)x)> (frac1ln a x-frac1ln(a+1)x )x (1+frac1ln ax+frac2ln^2 ax )-frac0.334xln^3 (a+1)x >fracln(1+1/a)ln^2 (a+1)x x-frac0.334 xln^3 (a+1)x$

multiplying by $ ln^2(a+1)x$ to get $ ln(1+frac1a) x > frac0.334xln a x => ln(1+1/a) > frac0.334ln a x => frac1a+0.5 > frac0.334ln ax $

giving $ ln a x > 0.334(a+0.5) $ exponentiation both sides gives $ a x > e^0.334 a+ 0.167 => x > frac1.2 e^0.334a a$

So for every real number $a geq 1$ the inequality is valid for all $ x geq max(3*10^11,frac1.2 e^0.334a a)$