Proof of Lemma 6

For each case, it is easy to check, by substitution, that conditions (2.14)-(2.16) are satisfied. Below, we prove condition (2.13) and $z>0$.

We begin with the first case, where $m_2^G = 2$. It is easy to see (2.13) is true if $z>0$, since

$$m_2^X = 2(1+z) > 2 > \frac{N+2}{N+1}.$$

Further, $z>0$ if $2\frac{N+3}{N+2} < m_3^G < 3$, which is true by $G\in \cal{\cal L}_N$, $r^G<\frac{3}{2}$, and $m_2^G = 2$.

Below, we consider the second case, where $m_2^G \neq 2$. We first prove $z>0$ by showing that $z$ is the larger solution of the two solutions of a quadratic equation that has a unique positive solution. Observe that

$$\begin{eqnarray*} % latex2html id marker 25396&\hspace{-4mm} & m_3^G(m_2^X+2z+... ...2^G(1+z) = \frac{N+3}{N+2}\left(m_2^G(1+z)-2z\right)^2 + 3zm_2^G \end{eqnarray*}$$

Thus, $z$ is a solution of the following quadratic equation: $f(z)=0$, where

$$f(z) \equiv \frac{N+3}{N+2}(m_2^G-2)^2z^2 - m_2^G\left((m_3... ...2^G-2)\right)z - (m_2^G)^2\left(r^G - \frac{N+3}{N+2}\right).$$

Since the coefficient of the leading term, $\frac{N+3}{N+2}(m_2^G-2)^2$, is positive and $f(0)<0$, there exists a unique positive solution of $f(z)=0$.

Second, we show $m_2^X\geq\frac{N+2}{N+1}$. We consider two cases: (i) $m_2^G\geq 2$ and (ii) $m_2^G< 2$. Case (i) is easy to show. Suppose $m_2^G\geq 2$. Observe that by (2.15),

$$m_2^X = z\left((m_2^G-2)z+2(m_2^G-1)\right) + m_2^G.$$

Thus, if $m_2^G\geq 2$, then $m_2^X\geq m_2^G \geq \frac{N+2}{N+1}$. Below, we consider case (ii).

Suppose $m_2^G< 2$. Observe that

$$m_2^X = -(2-m_2^G)z^2 + 2(m_2^G-1)z + m_2^G,$$

again by (2.15). Thus, $m_2^X\geq\frac{N+2}{N+1}$ iff $0 < z \leq z^*$, where $z^*$ is a larger solution, $x$, of the following quadratic equation: $\chi(x)=0$, where

$$\chi(x) = -(2-m_2^G)x^2 + 2(m_2^G-1)x + m_2^G - \frac{N+2}{N+1}.$$

That is,

$$z^* \equiv \frac{m_2^G-1 + \sqrt{\frac{N+2}{N+1}m_2^G-\frac{N+3}{N+1}}}{2-m_2^G}.$$ (A.1)

Thus, it suffices to show $f(z^*)\geq 0$. Since $\chi(z^*)=0$, we obtain $(z^*)^2$ as a linear function of $z^*$:

$$(z^*)^2 = \frac{2(m_2^G-1)z^* + m_2^G - \frac{N+2}{N+1}}{2-m_2^G}.$$

By substituting this $(z^*)^2$ into the expression for $f(z^*)$, we obtain

$$\begin{eqnarray*} \hspace{-6mm}f(z^*) & = & \frac{N+3}{N+2}(2-m_2^G)^2\left(\fra... ...N+2)}{2(N+1)(N+2)}\left((N+1)(m_2^G)^2+(2N+6)m_2^G-4(N+3)\right) \end{eqnarray*}$$

where the inequality follows from the assumption on $r^G$ in the lemma. By substituting (A.1) into the last expression, we obtain

$$f(z^*) = g(m_2^G) + h(m_2^G)\sqrt{\frac{(N+2)m_2^G-(N+3)}{N+1}},$$

where

$$\begin{eqnarray*} g(m_2^G) & \equiv & \frac{(N+1)^2(m_2^G)^2 - (N-3)(N+2)m_2^G -... ...& \equiv & \frac{(N+1)(m_2^G)^2 + 3(N+2)m_2^G - 4(N+3)}{2(N+2)}. \end{eqnarray*}$$

Since

$$g'(m_2^G) = 2(N+1)^2\left(m_2^G-\frac{N+2}{N+1}\right) + (N+2)(N+5) > 0$$

for $\frac{N+2}{N+1}\leq m_2^G<2$, $g(m_2^G)$ and $h(m_2^G)$ are increasing functions of $m_2^G$ in the range of $\frac{N+2}{N+1}\leq m_2^G<2$. Since

$$g(\frac{N+2}{N+1}) = \frac{2}{(N+1)^2(N+2)} > 0 \quad\mbox{and}\quad h(\frac{N+2}{N+1}) = \frac{2}{N^2+3N+2} > 0,$$

we have $g(m_2^G)\geq 0$ and $h(m_2^G)\geq 0$ for $\frac{N+2}{N+1}\leq m_2^G<2$. This implies $f(z^*)\geq 0$.     width 1ex height 1ex depth 0pt