JournalsrsmupVol. 147pp. 237–251

Inequalities involving π(x)\pi(x)

  • Horst Alzer

    Waldbröl, Germany
  • Man Kam Kwong

    The Hong Kong Polytechnic University, Hong Kong
  • József Sándor

    Babes-Bolyai University, Cluj-Napoca, Romania
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Abstract

We present several inequalities involving the prime-counting function π(x)\pi(x). Here, we give two examples of our results. We show that

169π(x)π(y)π2(x+y)\frac{16}{9} \pi(x)\pi(y)\leq \pi^2(x+y)

is valid for all integers x,y2x,y\geq 2. The constant factor 16/916/9 is the best possible. The special case x=yx=y leads to

43π(2x)π(x)(x=2,3,),\frac{4}{3}\leq \frac{\pi(2x)}{\pi(x)} \quad (x=2,3,\ldots ),

where the lower bound 4/34/3 is sharp. This complements Landau’s well-known inequality

π(2x)π(x)2(x=2,3,).\frac{\pi(2x)}{\pi(x)}\leq 2 \quad (x=2,3,\ldots ).

Moreover, we prove that the inequality

(2π(x+y)x+y)s(π(x)x)s+(π(y)y)s(0<sR)\Bigl(2\frac{\pi(x+y)}{x+y}\Bigr)^s \leq \Bigl( \frac{\pi(x)}{x}\Bigr)^s + \Bigl(\frac{\pi(y)}{y}\Bigr)^s \quad (0< s\in \mathbb{R})

holds for all integers x,y2x,y\geq 2 if and only if ss0=0.94745.s\leq s_0=0.94745\ldots. Here, s0s_0 is the only positive solution of

(167)t(65)t=1.\Bigl( \frac{16}{7}\Bigr)^t -\Bigl( \frac{6}{5}\Bigr)^t=1.

Cite this article

Horst Alzer, Man Kam Kwong, József Sándor, Inequalities involving π(x)\pi(x). Rend. Sem. Mat. Univ. Padova 147 (2022), pp. 237–251

DOI 10.4171/RSMUP/98