Rearrange the inequalities I have the two inequalities:
$$
f\left(\frac{x}{3}\right)-f\left(\frac{x}{4}\right)\le Ax+B\ln x-C
$$
$$
f\left(\frac{x}{3}\right)-f\left(\frac{x}{4}\right)+f\left(\frac{x}{6}\right)\ge Dx-B\ln x+E -\frac{F}{x}
$$
Where $f(x)$ is my function and $A, B, C, D, E$ are constants. My question is: Is it possible to rearrange these two inequalities, I don't know, combine them, add, substract, to obtain $$ f(x)<\ldots?$$ I tried many things, but I wasn't able to obtain the desired result. Thank you in advance! Can someone help me?
 A: This looks like
the inequalities Chebychev
came up work
to prove his
approximation to the
prime number theorem.
He proved that
there are constants $a$ and $b$
such that
$a\frac{n}{\ln n}
< \pi(n)
< b\frac{n}{\ln n}
$
starting with inequalities of these type.
I think he showed that
$a=0.95$ and $b=1.05$ worked.
Since other answers have been posted
since I started entering this,
I'll stop here
to avoid duplicating their answers.
So this is a comment
entered as an answer.
A: $$Ax + B\ln(x) - C\geq f(x/3) -f(x/4)$$
$$f(x/3) -f(x/4) + f(x/6) \geq Dx - B\ln(x) + E - \frac Fx$$
So
$$\left[Ax + B\ln x - C\right] + f(x/6) \geq Dx - B\ln(x) + E - \frac Fx$$
$$f(x/6) \geq \left(D-A\right)x - 2B\ln(x) + E + C - \frac F{x}$$
$$f(x) \geq 6(D-A)x - 2B\ln(x) - \frac F{6x} + E + C - 2B\ln(6)$$
A: Adding equation $(1)$ and $(2)$,
$$f\left(\frac{x}{3}\right)-f\left(\frac{x}{4}\right)+Dx-B\ln x+E -\frac{F}{x}\le f\left(\frac{x}{3}\right)-f\left(\frac{x}{4}\right)+f\left(\frac{x}{6}\right)+Ax+B\ln x-C$$
Rearranging, 
$$f(\frac x6)\ge (D-A)x-2B\ln x+(E+C)-\frac F x$$
Substituting $x\to \frac x6$,
$$f(x)\ge 6(D-A)x-2B\ln 6x+(E+C)-\frac F {6x}$$
