文摘
We establish an inequality by quadratic estimations; the double inequality $$ \frac{\pi^{2} x}{4 +\sqrt{(\pi^{2} -4)^{2} + (2\pi x)^{2}}} < \arctan{x} < \frac{\pi^{2} x}{4 +\sqrt{32+ (2\pi x)^{2}}} $$ holds for \(x>0\), where the constants \((\pi^{2} -4)^{2}\) and 32 are the best possible.KeywordsShafer’s inequalityan upper bound for arctangenta lower bound for arctangentMSC26D1542A101 IntroductionShafer [1–3] showed that the inequality $$ \arctan{x} > \frac{8 x}{3 +\sqrt{25+ \frac{80}{3} x^{2}}} $$ (1.1) holds for \(x>0\). Various Shafer-type inequalities are known, and they have been applied, extended and refined, see [4–8] and [9–12]. Especially, Zhu [12] showed an upper bound for inequality (1.1) and proved that the following double inequality $$ \frac{8 x}{3 +\sqrt{25 + \frac{80}{3}x^{2}}} < \arctan{x} < \frac{8 x}{3 +\sqrt{25+ \frac{256}{\pi^{2}} x^{2}}} $$ (1.2) holds for \(x>0\), where the constants \(80/3\) and \(256/\pi^{2}\) are the best possible. Recently, in [8], Sun and Chen proved that the following inequality $$ \arctan{x}< \frac{8 x+\frac{32}{4725}x^{7}}{3 +\sqrt{25+ \frac{80}{3} x^{2}}} $$ (1.3) holds for \(x>0\); moreover, they showed that the inequality $$ \frac{8 x+\frac{32}{4725}x^{7}}{3 +\sqrt{25+ \frac{80}{3} x^{2}}} < \frac{8 x}{3 +\sqrt{25+ \frac{256}{\pi^{2}} x^{2}}} $$ (1.4) holds for \(0< x< x_{0} \cong 1.4243\). In this paper, we shall establish the refinements of inequalities (1.2) and (1.3).2 Results and discussionMotivated by (1.2), (1.3) and (1.4), in this paper, we give inequalities involving arctangent. The following are our main results.