Some results of Young-type inequalities (Q2189982)

Young-type inequalities are extensions of the classical Young inequality \[a^{v}b^{1-v}\leq va+(1-v)b,\quad 0\leq v\leq 1,\ a, b>0.\] \textit{F. Kittaneh} and \textit{Y. Manasrah} [J. Math. Anal. Appl. 361, No. 1, 262--269 (2010; Zbl 1180.15021)] proved that \[ a^{v}+b^{1-v}+\min\{v, 1-v\}(\sqrt{a}-\sqrt{b})^2 \leq va+(1-v)b, \quad a, b>0,\ 0\leq v\leq 1. \] \textit{J. Zhao} and \textit{J. Wu} [J. Math. Anal. Appl. 421, No. 2, 1779--1789 (2015; Zbl 1298.15029)] gave further improvements of the above inequality and the reversed inequality in segment form. \textit{L. Zhu} [Rev. R. Acad. Cienc. Exactas Fís. Nat., Ser. A Mat., RACSAM 113, No. 2, 909--915 (2019; Zbl 1423.26051); Rev. R. Acad. Cienc. Exactas Fís. Nat., Ser. A Mat., RACSAM 114, No. 1, Paper No. 24, 11 p. (2020; Zbl 1434.26068)] presented new refinements of Young's inequality. In this paper, some improvements of Young-type inequality under some conditions are presented for the geometric mean \(a\sharp_\nu b:= a^{1-\nu}b^\nu\) and the arithmetic mean \(a\nabla_{\nu} b:= (1-\nu) a+\nu b\) of two positive numbers \(a, b\). Theorem. Let \(\nu, \tau\) and \(a, b\) be real positive numbers with \(0<\nu, \tau<1.\) Then \[\frac{a\nabla_{\nu} b-a\sharp_\nu b}{a\nabla_{\tau} b-a\sharp_\tau b}\leq \frac{\nu(1-\nu)}{\tau(1-\tau)},\quad \mbox{for} \ (b-a)(\nu-\tau)\geq 0,\] \[\frac{a\nabla_{\nu} b-a\sharp_\nu b}{a\nabla_{\tau} b-a\sharp_\tau b}\geq \frac{\nu(1-\nu)}{\tau(1-\tau)},\quad \mbox{for}\ (b-a)(\nu-\tau)\leq 0.\] The result provides improvements of Young-type inequalities of \textit{H. Alzer} et al. [Linear Multilinear Algebra 63, No. 3, 622--635 (2015; Zbl 1316.15023)]. Theorem. Let \(\nu, \tau\) and \(a, b\) be real positive numbers with \(0<\nu, \tau<1.\) Then \[\frac{(a\nabla_{\nu} b)^2-(a\sharp_\nu b)^2}{(a\nabla_{\tau} b)^2-(a\sharp_\tau b)^2}\leq \frac{\nu(1-\nu)}{\tau(1-\tau)},\quad \mbox{for}\ (b-a)(\nu-\tau)\geq 0,\] \[\frac{(a\nabla_{\nu} b)^2-(a\sharp_\nu b)^2}{(a\nabla_{\tau} b)^2-(a\sharp_\tau b)^2}\geq \frac{\nu(1-\nu)}{\tau(1-\tau)},\quad \mbox{for}\ (b-a)(\nu-\tau)\leq 0.\] The author also obtains some related results about operators and Hilbert-Schmidt norms and Kantorovich constant.