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

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

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})$
