1. Introduction Let r∈(0,1) and a,b>0. Then the elliptic elliptic integral of the first kind K(r) and second kind E(r), Gaussian arithmetic-geometric mean AG(a,b), arithmetic mean A(a,b), and quadratic mean Q(a,b) are, respectively, given by (1)Kr=∫0π/211-r2sin2tdt,Er=∫0π/21-r2sin2t dt,(2)AGa,b=π2∫0π/2dt/a2cos2t+b2sin2t,Aa,b=a+b2,Qa,b=a2+b22.

The Gauss identity [1–3] shows that (3)AG1,r′=π2Krfor all r∈(0,1), where and in what follows r′=1-r2.

It is well known that the elliptic elliptic integrals K(r) and E(r) and the Gaussian arithmetic-geometric mean AG(a,b) have many applications in mathematics, physics, mechanics, and engineering [4–9]. Recently, the bounds for the Gaussian arithmetic-geometric mean AG(a,b) have attracted the attention of many researchers.

The inequalities (4)1+r2AG1,r<AG1,r<π2log4/r,(5)La,b<AGa,b<L3/2a,bfor all r∈(0,1) and a,b>0 with a≠b can be found in the literature [10–12], where L(a,b)=(a-b)/(loga-logb) and Lp(a,b)=L1/pap,bp are, respectively, the logarithmic and pth generalized logarithmic means of a and b. The first inequality of (5) is due to Carlson and Vuorinen [13].

By using a variant of L’Hospital’s rule and representation theorems with elliptic integrals, Vamanamurthy and Vuorinen [14] proved, among other results, the inequalities (6)AGa,b<Aa,bLa,b,(7)La,b<AGa,b<π2La,b,(8)AGa,b<Ia,b<Aa,b,(9)AGa,b<Aa,b+Ga,b2,(10)Aa,b<AGa2,b2AGa,b<Qa,b,(11)AGa,b>L1/λaλ,bλfor all a,b>0 with a≠b and λ∈(0,1], where I(a,b)=(bb/aa)1/(b-a)/e is the identric mean of a and b.

By use of the homogeneity of the above means and a series representation of AG(a,b) due to Gauss, Sándor [15] obtained, among other results, new proofs for inequalities (7), (8) and a counterpart of inequality (9): (12)AGa,b>Aa,bGa,b,for all a,b>0 with a≠b, where G(a,b)=ab is the geometric mean of a and b. Inequalities (9) and (12) show that AG lies between the arithmetic and geometric means of A and G. In [16], Sándor provided new proofs for inequalities (6) and (8), (9), (10), and (12) by using only elementary methods for recurrent sequences and found much stronger forms of these results.

Neuman and Sándor [17] gave the comparison of the Gaussian arithmetic-geometric mean and the Schwab-Borchardt mean.

The upper bounds π/2log(4/r) for AG(1,r) in (4) were replaced by π(1-r2/9)/[2log(4/r)] due to Kühnau [18].

Qiu and Vamanamurthy [19] presented that 4π/[(9-r2)(2log2-logr)] and (9-r2)π/[18.192×(2log2-logr)] are, respectively, the lower and upper bounds for AG(1,r) with r∈(0,1). Alzer and Qiu [20] proved that λ=3/4 and μ=2/π are the best possible parameters such that the double inequality (13)1λ/La,b+1-λ/Aa,b<AGa,b<1μ/La,b+1-μ/Aa,bholds for all a,b>0 with a≠b.

Chu and Wang [21] proved that the double inequality (14)Spa,b<AGa,b<Sqa,bholds for all a,b>0 with a≠b if and only if p≤1/2 and q≥1, where Sp(a,b)=[ap-1+bp-1/(a+b)]1/(p-2) (p≠2) and S2(a,b)=aabb1/(a+b) is the pth Gini mean of a and b. In [22], Yang et al. proved that the inequalities(15)S7/4,-1/4a,b<AGa,b<A1/4a,bL3/4a,b,AGa,b<Sp,1a,bS1-p,1a,bhold for all p∈(1/2,1) and a,b>0 with a≠b, where Sp,q(a,b)=[q(ap-bp)/(p(aq-bq))]1/(p-q) is the Stolarsky mean [23] of a and b.

Let a,b>0 with a≠b and x∈[1/2,1]. Then it is not difficult to verify that the function f(x)=Q[xa+(1-x)b,xb+(1-x)a] is continuous and strictly increasing on the interval [1/2,1]. Note that(16)f12=Aa,b=minAa,b,Qa,b<AGAa,b,Qa,b,(17)AGAa,b,Qa,b<maxAa,b,Qa,b=Qa,b=f1.

Inequalities (16) give us the motivation to deal with the best possible parameters α1,β1,α2,β2∈R and α3,β3∈(1/2,1) such that the double inequalities (18)Qα1a,bA1-α1a,b<AGAa,b,Qa,b<Qβ1a,bA1-β1a,b,α2Qa,b+1-α2Aa,b<AGAa,b,Qa,b<β2Qa,b+1-β2Aa,b,Qα3a+1-α3b,α3b+1-α3a<AGAa,b,Qa,b<Qβ3a+1-β3b,β3b+1-β3ahold for all a,b>0 with a≠b.

3. Main Results Theorem 4. The double inequality (24)Qα1a,bA1-α1a,b<AGAa,b,Qa,b<Qβ1a,bA1-β1a,bholds for all a,b>0 with a≠b if and only if α1≤1/2 and β1≥2[logπ-logK(2/2)]/log2-1=0.5215….

Proof. Since A(a,b), Q(a,b), and AG(a,b) are symmetric and homogenous of degree 1, without loss of generality, we assume that a>b>0. Let r=(a-b)/2(a2+b2)∈(0,2/2). Then (2) and (3) lead to (25)AGAa,b,Qa,b=πAa,b2r′Kr,Qa,b=Aa,br′,(26)logAGAa,b,Qa,b-logAa,blogQa,b-logAa,b=logKr+logr′+log2-logπlogr′.Let (27)f1r=logKr+logr′+log2-logπ,f2r=logr′,fr=f1rf2r.Then simple computations give (28)f10+=f20+=0,f1′rf2′r=Kr-Err2Kr.

It follows from Lemmas 1, 2(3) and (27) and (28) that (29)f0+=12and f(r) is strictly increasing on the interval (0,2/2).

Note that (30)f22-=2logπ-logK2/2log2-1.

Therefore, Theorem 4 follows easily from (26), (27), (29), and (30) and the monotonicity of f(r) on the interval (0,2/2).

Remark 5. The left side inequality of Theorem 4 for α1≤1/2 can be derived directly from the fact that AG(a,b)>G(a,b)=ab and Q(a,b)>A(a,b) for all a,b>0 with a≠b.

Theorem 6. The double inequality (31)α2Qa,b+1-α2Aa,b<AGAa,b,Qa,b<β2Qa,b+1-β2Aa,bholds for all a,b>0 with a≠b if and only if α2≤[π-2K(2/2)]/[(2-2)K(2/2)]=0.4783… and β2≥1/2.

Proof. Without loss of generality, we assume that a>b>0. Let r=(a-b)/2(a2+b2)∈(0,2/2). Then it follows from (2) and (3) that (32)AGAa,b,Qa,b-Aa,bQa,b-Aa,b=π/2Kr-r′1-r′.Let (33)g1r=π2Kr-r′,g2r=1-r′,gr=g1rg2r.

Then simple computations lead to (34)g10+=g20+=0,g1′rg2′r=1-π2Er-r′2Krr2r′1/2Kr-2.

It follows from Lemmas 1, 2(1) and (2) together with (33) and (34) that (35)g0+=12and g(r) is strictly decreasing on the interval (0,2/2).

Note that (36)g22-=π-2K2/22-2K2/2.

Therefore, Theorem 6 follows easily from (32), (33), (35), and (36) and the monotonicity of g(r) on the interval (0,2/2).

Remark 7. The right side inequality of Theorem 6 for β2≥1/2 can be derived directly from the fact that AG(a,b)<A(a,b)=(a+b)/2 and Q(a,b)>A(a,b) for all a,b>0 with a≠b.

Theorem 8. Let α3,β3∈(1/2,1). Then the double inequality (37)Qα3a+1-α3b,α3b+1-α3a<AGAa,b,Qa,b<Qβ3a+1-β3b,β3b+1-β3aholds for all a,b>0 with a≠b if and only if α3≤1/2+2π2-2K2(2/2)/[4K(2/2)]=0.8299… and β3≥1/2+2/4=0.8535….

Proof. Without loss of generality, we assume that a>b>0. Let r=(a-b)/2(a2+b2)∈(0,2/2) and p∈(1/2,1). Then (2) and (3) lead to (38)Qpa+1-pb,pb+1-pa=1-4p1-pr2r′Aa,b,(39)Qpa+1-pb,pb+1-pa-AGAa,b,Qa,b=Aa,b1-4p1-pr2+π/2Krr′Hr,where(40)Hr=1-4p1-pr2-π24K2r,(41)H0+=0,(42)H22-=1-2p1-p-π24K22/2,(43)H′r=π2r2hr,where h(r) is defined by (21).

We divide the proof into four cases.

Case 1 (p=p0=1/2+2π2-2K2(2/2)/[4K(2/2)]). Then (42) becomes(44)H22-=0.

It follows from Lemma 3 and (43) that there exists r0∈(0,2/2) such that H(r) is strictly decreasing on (0,r0] and strictly increasing on [r0,2/2). Therefore, (45)Qp0a+1-p0b,p0b+1-p0a<AGAa,b,Qa,bfollows from (39), (41), and (44) together with the piecewise monotonicity of H(r) on the interval (0,2/2).

Case 2 (p=p0∗=1/2+2/4). Then we clearly see that (46)Qp0∗a+1-p0∗b,p0∗b+1-p0∗a=A2a,b+Q2a,b2=QAa,b,Qa,b>AAa,b,Qa,b>AGAa,b,Qa,b.

Case 3 (1/2+2π2-2K2(2/2)/[4K(2/2)]<p<1). Then (42) leads to (47)H22->0.

Equation (39) and inequality (47) imply that there exists small enough 0<δ1<2/2 such that (48)Qpa+1-pb,pb+1-pa>AGAa,b,Qa,bfor all a,b>0 with |a-b|/2(a2+b2)∈(2/2-δ1,2/2).

Case 4 (1/2<p<1/2+2/4). Then (40) leads to (49)Hr=4p-122-12r2+or2.Note that (50)4p-122-12r2<0.

Equations (39) and (49) together with inequality (50) imply that there exists small enough 0<δ2<2/2 such that (51)Qpa+1-pb,pb+1-pa<AGAa,b,Qa,bfor all a,b>0 with a-b/2(a2+b2)∈(0,δ2).