Sharp Bounds for the Weighted Geometric Mean of the First Seiffert and Logarithmic Means in terms of Weighted Generalized Heronian Mean

Theorem 1. Let a, b > 0, a ≠ b, α ∈ (0,1). Then

where p = p(α) = +∞, q(α) = 2(2 − α)/(1 + α) are the best possible functions.

Proof. First, we prove the left inequality of (8). The inequalities (1) imply that

From $$ for α ∈ (0,1) (see (14)) we obtain that p(α) = +∞ is the optimal function.

Without loss of generality, we assume that 0 < a < b. Let $$; then 0 < t < 1. The right inequality of (8) can be rewritten as

Simple computations lead to Then the inequality (11) is equivalent to Denote From r(1) = 0 and r^′(t) = (2 − 4t + 2t²)/(t + t³) > 0 we have r(t) < 0 for t ∈ (0,1). It implies ((π − 4arctant)/(−2ln t)) ^α < 1. From v(1) = 0, v^′(1) = 0, v^′′(t) = 2 − 2/t < 0 we obtain v^′(t) > 0 and so v(t) < 0. It implies that s(t, α) > 0 for t, α∈(0,1). This leads to If we show $$ for t, α ∈ (0,1), then $$ will be the best function in (8). Simple computations lead to $$ which is equivalent to Using the inequality t^α < 1 − α(1 − t) for t, α ∈ (0,1) it suffices to show that It will be done, if we show R(t, 0) < 0 and R(t, 1) < 0. It follows from R(t, α) being a linear continuous function in the argument α is equivalent to From s(1) = 0 it suffices to show that s^′(t) > 0 which is equivalent to It follows from v(1) = 0 and where w(t) = −(1 − t) ²(1 − t²) ².

Next, we show that

The inequality (21) is equivalent to So, it suffices to show that It is easy to see that Because of it suffices to prove g₁(t) > 0 for 0 < t < 1. From arctan(t) > t − t³/3, for t ∈ (0,1), g₂(1) = 0 we have done it, if we show on (0,0.67〉 and $$ on 〈0.67,1).

Simple computation gives

The inequality $$ is equivalent to From ch^′′′(t) < 0 we get ch^′′(t) is a decreasing function. ch^′′(0.67) = −324.3366 implies ch^′′(t) < 0 on 〈0.67,1). So, we obtain ch^′(t) is a decreasing function. From ch^′(0.67) = −54.8414 we have ch^′(t) < 0 on 〈0.67,1). It implies that ch(t) is a decreasing function. From ch(0.67) = −1.9053 we get ch(t) < 0 on 〈0.67,1). So $$ on 〈0.67,1).

Next, we show g₃(t) > 0 on (0,0.67〉.

Simple computation gives

The inequality g₃(t) > 0 is equivalent to on (0,0.67〉. From h(0) = 16 − 5π > 0, h(0.1) = 0.1453, h(0.15) = 0.1427, h(0.67) = 0.2602, h^′(0.15) = 0.3998 it suffices to show that h^′(t) < 0 on (0,0.1〉; h(t) > 0 on 〈0.1,0.15〉 and h^′(t) has only one root in (0.15,0.67).

First, we show h(t) > 0 on 〈0.1,0.15〉. From t³ > 0.1t², t⁴ < 0.15²t², t⁵ < 0.15³t² we have

where It is easy to see that l^′(t) = 0 for t = (10π − 29)/18.6 = 0.1299. From l^′′(t) > 0 on 〈0.1,0.15〉 and l(0.1299) = 0.1351 we have l(t) > 0. It implies h(t) > 0 on 〈0.1,0.15〉.

Next, we show h^′(t) < 0 on (0,0.1〉. Simple computation gives

where j(0) = 29 − 10π < 0, j^′′(t) > 0, j(0.1) = −0.45750 imply j(t) < 0 so h^′(t) < 0 on (0,0.1〉.

Finally, we show that h^′(t) has only one root on (0.15,0.67). From h^′′′′(t) < 0 we obtain h^′′′(t) is a decreasing function. Because of h^′′′(0.15) = −37 we have h^′′′(t) < 0 on (0.15,0.67) so h^′(t) is a concave function. From h^′(0.15) = 0.3998 and h^′(0.67) = −8.2333 we have that h^′(t) has only one root on (0.15,0.67). It implies h(t) > 0 on 〈0.15,0.67〉. So, the proof of decreasing of G(t, α) is complete.

In what follows, we find the representation of the function q(α).

It is easy to see that

Equation (36) can be rewritten as where Simple computations give where y(t) is a suitable function. Similarly we have where u(t) is a suitable function. Denote S(t, α) = Y(t, α)U(t, α). Then where s(t) is a suitable function. Using the L’Hospital′s rule we obtain The proof is complete.