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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?

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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.

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  • $\begingroup$ The answers doesn't provide the solution I was looking for, despite the effort. $\endgroup$ – VanDerWarden Aug 25 '15 at 16:05
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$$Ax + B\ln(x) - C\geq f(x/3) -f(x/4)$$ $$f(x/3) -f(x/4) + f(x/6) \geq Dx - B\ln(x) + E - \frac Fx$$ So $$\left[Ax + B\ln x - C\right] + f(x/6) \geq Dx - B\ln(x) + E - \frac Fx$$ $$f(x/6) \geq \left(D-A\right)x - 2B\ln(x) + E + C - \frac F{x}$$ $$f(x) \geq 6(D-A)x - 2B\ln(x) - \frac F{6x} + E + C - 2B\ln(6)$$

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  • $\begingroup$ Note that I was looking for $f(x)<\ldots$... $\endgroup$ – VanDerWarden Aug 25 '15 at 15:56
  • $\begingroup$ I'm not sure you can achieve that with the information provided $\endgroup$ – jameselmore Aug 25 '15 at 15:57
  • $\begingroup$ I remember somewhere was the example that if $ f(x)-f\left(\frac{x}{2}\right)<\frac{3x}{4} $ and $ f(x)-f\left(\frac{x}{2}\right)+f\left(\frac{x}{3}\right)>\frac{2x}{3} $, then (after "changing $ x $ to $ \frac{x}{2} $, $ \frac{x}{4} $, $ \ldots $ and adding the results" - I don't understand this method), $ f(x)<\frac{3x}{2} $. $\endgroup$ – VanDerWarden Aug 25 '15 at 16:03
  • $\begingroup$ Use your expression for $f(x)$ to bound the $f(x/4)$ term in (1) and you're done. $\endgroup$ – Macavity Aug 25 '15 at 16:18
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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 $x\to \frac x6$,

$$f(x)\ge 6(D-A)x-2B\ln 6x+(E+C)-\frac F {6x}$$

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  • $\begingroup$ Note that I was looking for $ f(x)<\ldots$... $\endgroup$ – VanDerWarden Aug 25 '15 at 15:55
  • $\begingroup$ Well this is what I can conclude. Sorry if it doesn't help... $\endgroup$ – Mythomorphic Aug 25 '15 at 15:56
  • $\begingroup$ Use your expression for $f(x)$ to bound the $f(x/4)$ term in (1) and you're done. $\endgroup$ – Macavity Aug 25 '15 at 16:17
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I was able to compute the upper bound. I posted the answer in a new question, see Upper bound of the function.

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