The calculation in the proof is purely elementary. The order estimate involving the parameter $Q$ will be used at crucial places for our later estimates, so we include the detailed proof. Plugging the critical points formulae (5.114) and (5.115) into the expression of $F$ and $G$,

(5.140)

$$F(t_{0})+G(u_{0})=yt_{0}+(-y)^{\frac{1}{2}}u_{0}-\frac{2Q+\gamma_{n}}{2}+Q\log\frac{t_{0}}{t_{0}+1}+(2Q+\gamma_{n})\log u_{0},$$

where $Q\equiv\alpha-\beta-1$ and $\gamma_{n}\equiv\frac{1}{2}+\frac{1}{n}$ as before.

First, it is straightforward that

(5.141)

$$yt_{0}+(-y)^{\frac{1}{2}}u_{0}\leq(-y)+\frac{\gamma_{n}}{2}.$$

So this implies that

(5.142)		$\displaystyle e^{F(t_{0})+G(u_{0})}$	$\displaystyle\leq$	$\displaystyle C_{n}\cdot e^{-y}\cdot e^{-Q}\cdot\Big(\frac{t_{0}}{1+t_{0}}\Big)^{Q}\cdot u_{0}^{2Q+\gamma_{n}}$
			$\displaystyle=$	$\displaystyle C_{n}\cdot e^{-y}\cdot e^{-Q}\cdot\Big(\frac{t_{0}^{2}}{t_{0}(1+t_{0})}\Big)^{Q}\cdot u_{0}^{2Q+\gamma_{n}}$
			$\displaystyle=$	$\displaystyle C_{n}\cdot e^{-y}\cdot u_{0}^{\gamma_{n}}\cdot e^{-Q}\cdot\frac{(u_{0}t_{0})^{2Q}}{(\frac{Q}{-y})^{Q}},$

where the last equality follows from (5.112).

Now we claim

(5.143)

$$u_{0}t_{0}\leq(-y)^{-\frac{1}{2}}\cdot(Q+\frac{\gamma_{n}}{2}).$$

To prove this, we denote $\tau\equiv\frac{2\gamma_{n}}{-y}>0$ and $\widehat{Q}\equiv\frac{4Q}{-y}>0$. Then using the critical point formulae of $u_{0}$ and $t_{0}$ given by (5.114) and (5.115), we obtain

(5.144)				$\displaystyle u_{0}t_{0}$
			$\displaystyle=$	$\displaystyle\frac{(-y)^{\frac{1}{2}}}{4}\cdot\Big(1+\sqrt{1+\widehat{Q}}\Big)\cdot\Big(-1+\sqrt{1+\widehat{Q}+\tau}\Big)$
			$\displaystyle=$	$\displaystyle\frac{(-y)^{\frac{1}{2}}}{4}\cdot\Big(-1+\sqrt{1+\widehat{Q}}\cdot\sqrt{1+\widehat{Q}+\tau}+\sqrt{1+\widehat{Q}}-\sqrt{1+\widehat{Q}+\tau}\Big)$
			$\displaystyle\leq$	$\displaystyle\frac{(-y)^{\frac{1}{2}}}{4}\cdot\Big(-1+\sqrt{1+\widehat{Q}+\tau}\cdot\sqrt{1+\widehat{Q}+\tau}+\sqrt{1+\widehat{Q}+\tau}-\sqrt{1+\widehat{Q}+\tau}\Big)$
			$\displaystyle=$	$\displaystyle\frac{(-y)^{\frac{1}{2}}}{4}\cdot(\widehat{Q}+\tau)$
			$\displaystyle=$	$\displaystyle(-y)^{-\frac{1}{2}}\cdot(Q+\frac{\gamma_{n}}{2}).$

Then it follows that

(5.145)

$$\frac{(u_{0}t_{0})^{2Q}}{(\frac{Q}{-y})^{Q}}\leq\frac{(Q+\frac{\gamma_{n}}{2})^{2Q}}{Q^{Q}}=Q^{Q}\cdot(1+\frac{\gamma_{n}}{2Q})^{2Q}\leq e^{\gamma_{n}}\cdot Q^{Q}.$$

Moreover, we notice that

(5.146)

$$u_{0}^{\gamma_{n}}\leq C_{n}\cdot Q^{\frac{\gamma_{n}}{2}}\cdot(-y)^{\frac{\gamma_{n}}{2}}.$$

Therefore, combining all the above, we have

(5.147)		$\displaystyle e^{F(t_{0})+G(u_{0})}$	$\displaystyle\leq$	$\displaystyle C_{n}\cdot e^{-y}\cdot u_{0}^{\gamma_{n}}\cdot e^{-Q}\cdot\frac{(u_{0}t_{0})^{2Q}}{(\frac{Q}{-y})^{Q}}$
			$\displaystyle\leq$	$\displaystyle C_{n}\cdot(-y)^{\frac{\gamma_{n}}{2}}\cdot e^{-y}\cdot e^{-Q}\cdot Q^{Q+\frac{\gamma_{n}}{2}}.$

∎