Abstract
In this paper, we prove that the double inequality a, b>0a\neq bp\leq 4\log 2/(4+2\log 2-π)=1.2351\cdotsq\geq 4/3% M_{r}(a,b)=[(a^{r}+b^{r})/2]^{1/r}(r\neq 0)M_{0}(a,b)=\sqrt{ab}rB(a,b)=Q(a,b)e^{A(a,b)/T(a,b)-1}A(a,b)=(a+b)/2Q(a,b)=\sqrt{(a^{2}+b^{2})/2}% T(a,b)=(a-b)/[2\arctan((a-b)/(a+b))]$.
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