Abstract
In the article, we prove that lambda _{1}=1/2+sqrt{ [ (sqrt{2}+ log (1+sqrt{2}) )/2 ]^{1/nu }-1}/2, mu _{1}=1/2+sqrt{6 nu }/(12nu ), lambda _{2}=1/2+sqrt{ [(pi +2)/4 ] ^{1/nu }-1}/2 and mu _{2}=1/2+sqrt{3nu }/(6nu ) are the best possible parameters on the interval [1/2, 1] such that the double inequalities \t\t\tCν[λ1x+(1−λ1)y,λ1y+(1−λ1)x]A1−ν(x,y)<RQA(x,y)<Cν[μ1x+(1−μ1)y,μ1y+(1−μ1)x]A1−ν(x,y),Cν[λ2x+(1−λ2)y,λ2y+(1−λ2)x]A1−ν(x,y)<RAQ(x,y)<Cν[μ2x+(1−μ2)y,μ2y+(1−μ2)x]A1−ν(x,y)\\documentclass[12pt]{minimal}\t\t\t\t\\usepackage{amsmath}\t\t\t\t\\usepackage{wasysym}\t\t\t\t\\usepackage{amsfonts}\t\t\t\t\\usepackage{amssymb}\t\t\t\t\\usepackage{amsbsy}\t\t\t\t\\usepackage{mathrsfs}\t\t\t\t\\usepackage{upgreek}\t\t\t\t\\setlength{\\oddsidemargin}{-69pt}\t\t\t\t\\begin{document} $$\\begin{aligned}& C^{\\nu }\\bigl[\\lambda _{1}x+(1-\\lambda _{1})y, \\lambda _{1}y+(1-\\lambda _{1})x\\bigr]A ^{1-\\nu }(x, y) \\\\& \\quad < \\mathcal{R}_{QA}(x, y)< C^{\\nu }\\bigl[\\mu _{1}x+(1-\\mu _{1})y, \\mu _{1}y+(1-\\mu _{1})x\\bigr]A^{1-\\nu }(x, y), \\\\& C^{\\nu }\\bigl[\\lambda _{2}x+(1-\\lambda _{2})y, \\lambda _{2}y+(1-\\lambda _{2})x\\bigr]A ^{1-\\nu }(x, y) \\\\& \\quad < \\mathcal{R}_{AQ}(x, y)< C^{\\nu }\\bigl[\\mu _{2}x+(1-\\mu _{2})y, \\mu _{2}y+(1-\\mu _{2})x\\bigr]A^{1-\\nu }(x, y) \\end{aligned}$$ \\end{document} hold for all x, y>0 with xneq y and nu in [1/2, infty ), where A(x, y) is the arithmetic mean, C(x, y) is the contraharmonic mean, and mathcal{R}_{QA}(x, y) and mathcal{R}_{AQ}(x, y) are two Neuman means.
Highlights
Lemma 2.3 The function t3 φ(t) = (1 + t2) arctan(t) – t is strictly increasing from (0, 1) onto (3/2, 2/(π – 2))
Lemma 2.4 Let θ ∈ [0, 1], ν ∈ [1/2, ∞), t ∈ (0, 1) and fθ,ν(t) = ν log 1 + θ t2 – log t 1 + t2 + sinh–1(t) + log t + log 2
Lemma 2.5 Let θ ∈ [0, 1], ν ∈ [1/2, ∞), t ∈ (0, 1) and gθ,ν(t) = ν log 1 + θt2 – log t + 1 + t2 arctan(t) + log(t) + log 2
Summary
Lemma 2.3 The function t3 φ(t) = (1 + t2) arctan(t) – t is strictly increasing from (0, 1) onto (3/2, 2/(π – 2)). Proof Let φ1(t) = t3 and φ2(t) = (1 + t2) arctan(t) – t.
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