**Lemma 10.** *We have*

*(i) If* 0 ≤ *ρ* < *ρ*ˆ1*, then*

$$\inf\_{(t,s)\in\overline{A}}\mathbb{g}(t,s) = \mathbb{g}\_A(t\_{A\prime}s\_A) = 4(\mu\_2 + (1 - 2\rho)\mu\_1)\_{\prime\prime}$$

*where* (*tA*,*sA*) ∈ *A is the unique minimizer of g*(*<sup>t</sup>*,*<sup>s</sup>*) *on A. (ii) If ρ* = *ρ*ˆ1*, then*

$$\inf\_{(t,s)\in\overline{A}}\mathbb{g}(t,s) = \mathbb{g}\_A(t\_{A'}s\_A) = 4(\mu\_2 + (1-2\rho)\mu\_1)\_{\prime\prime}$$

*where* (*tA*,*sA*)=(*t*<sup>∗</sup>,*s*<sup>∗</sup>) ∈ *L is the unique minimizer of g*(*<sup>t</sup>*,*<sup>s</sup>*) *on A.*

*(iii) If ρ*ˆ1 < *ρ* < *ρ*ˆ2*, then*

$$\inf\_{(t,s)\in\overline{A}}\mathcal{g}(t,s) = \mathcal{g}\_L(s^\*) = \frac{2}{1+\rho}(\mu\_1 + \mu\_2 + 2/s^\*)\_{\rho}$$

*where* (*s*<sup>∗</sup>,*s*<sup>∗</sup>) *is the unique minimizer of g*(*<sup>t</sup>*,*<sup>s</sup>*) *on A.(iv) If ρ* = *ρ*ˆ2*, then <sup>t</sup>*<sup>∗</sup>(*ρ*ˆ2) = *s*<sup>∗</sup>(*ρ*ˆ2) = 1/*μ*2 *and*

$$\inf\_{(t,s)\in\overline{A}}\lg(t,s) = \lg(s^\*) = \lg(1/\mu\_2) = 4\mu\_{2,1}$$

*where the minimum of g*(*<sup>t</sup>*,*<sup>s</sup>*) *on A is attained at* (1/*μ*2, 1/*μ*2)*, with g*3(1/*μ*2, 1/*μ*2) = *g*2(1/*μ*2)*. (v) If ρ*ˆ2 < *ρ* < 1*, then*

$$\inf\_{(t,s)\in\overline{A}}\mathcal{g}(t,s) = \mathcal{g}\_2(1/\mu\_2) = 4\mu\_{2\prime}$$

*where the minimum of g*(*<sup>t</sup>*,*<sup>s</sup>*) *on A is attained when g*(*<sup>t</sup>*,*<sup>s</sup>*) = *g*2(*s*) *on D*2 *(see Figure 1).*

Consequently, combining the results in Lemma 8 and Lemma 10, we conclude that Proposition 3 holds for 0 ≤ *ρ* < 1. Thus, the proof is complete.
