# 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?

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.

• The answers doesn't provide the solution I was looking for, despite the effort. – VanDerWarden Aug 25 '15 at 16:05

$$Ax + B\ln(x) - C\geq f(x/3) -f(x/4)$$ $$f(x/3) -f(x/4) + f(x/6) \geq Dx - B\ln(x) + E - \frac Fx$$ So $$\left[Ax + B\ln x - C\right] + f(x/6) \geq Dx - B\ln(x) + E - \frac Fx$$ $$f(x/6) \geq \left(D-A\right)x - 2B\ln(x) + E + C - \frac F{x}$$ $$f(x) \geq 6(D-A)x - 2B\ln(x) - \frac F{6x} + E + C - 2B\ln(6)$$

• Note that I was looking for $f(x)<\ldots$... – VanDerWarden Aug 25 '15 at 15:56
• I'm not sure you can achieve that with the information provided – jameselmore Aug 25 '15 at 15:57
• 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), $f(x)<\frac{3x}{2}$. – VanDerWarden Aug 25 '15 at 16:03
• Use your expression for $f(x)$ to bound the $f(x/4)$ term in (1) and you're done. – Macavity Aug 25 '15 at 16:18

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 $x\to \frac x6$,

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

• Note that I was looking for $f(x)<\ldots$... – VanDerWarden Aug 25 '15 at 15:55
• Well this is what I can conclude. Sorry if it doesn't help... – Mythomorphic Aug 25 '15 at 15:56
• Use your expression for $f(x)$ to bound the $f(x/4)$ term in (1) and you're done. – Macavity Aug 25 '15 at 16:17

I was able to compute the upper bound. I posted the answer in a new question, see Upper bound of the function.