LAUSR

In the article, we prove that the double inequalities $$\begin{array}{*{20}{c}} {{G^p}\left[ {{\lambda _1}a + \left( {1 - {\lambda _1}} \right)b,{\lambda _1}b + \left( {1 - {\lambda _1}} \right)a} \right]{A^{1 - p}}\left( {a,b} \right) < T\left[ {A\left( {a,b} \right),G\left( {a,b} \right)} \right]} \\ { < {G^p}\left[ {\mu _1^{}a + \left( {1 - {\mu _1}} \right)a} \right]{A^{1 - p}}\left( {a,b} \right),\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;} \\ {{C^s}\left[ {{\lambda _2}a + \left( {1 - {\lambda _2}} \right)b,{\lambda _2}b + \left( {1 - {\lambda _2}} \right)a} \right]{A^{1 - s}}\left( {a,b} \right) < T\left[ {A\left( {a,b} \right),Q\left( {a,b} \right)} \right]} \\ { < {C^s}\left[ {{\mu _2}a + \left( {1 - {\mu _2}} \right)b,{\mu _2}b + {{\left( {1 - {\mu _2}} \right)}_a}} \right]{A^{1 - p}}\left( {a,b} \right)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;} \end{array}$$ hold for all a, b > 0 with a ≠ b if and only if $${\lambda _1} \le 1/2 - \sqrt {1 - {{\left({2/\pi} \right)}^{2/p}}} /2$$ , $${\mu _1} \ge 1/2 - \sqrt {2p} /\left({4p} \right),{\lambda _2} \le 1/2 + \sqrt {{2^{3/\left({2s} \right)}}{{\left({\varepsilon \left({\sqrt 2 /2} \right)/\pi} \right)}^{1/s}} - 1} /2$$ and $${\mu _2} \ge 1/2 + \sqrt s /\left({4s} \right)$$ if λ1, μ1 ∈ (0, 1/2), λ2, μ2 ∈ (1/2, 1), p ≥ 1 and s ≥ 1/2, where $$G\left({a,b} \right) = \sqrt {ab} $$ , A(a, b) = (a + b)/2, $$T\left({a,b} \right) = 2\int_0^{\pi /2} {\sqrt {{a^2}{{\cos}^2}t + {b^2}{{\sin}^2}t} dt/\pi} $$ , $$Q\left({a,b} \right) = \sqrt {\left({{a^2} + {b^2}} \right)/2} $$ , C(a, b) = (a2 + b2)/(a + b) and $$\varepsilon (r) = \int_0^{\pi /2} {\sqrt {1 - {r^2}{{\sin}^2}t}} {\rm{d}}t$$ .