Define $$E=\frac{F}{4}n\lambda^2-\frac{F}{2}\lambda^2-\frac{K}{2}n\lambda-G\lambda+\frac{n}{4}K-\frac{1}{2}K$$ and $$P=-\frac{1}{8}Hn\lambda^3+\frac{1}{4}H\lambda^3-4H\lambda^2-\frac{1}{16}Hn\lambda^4+\frac{1}{8}H\lambda^4+\frac{3}{16}Hn-\frac{3}{8}H$$ where $$K=\sin\alpha+2\sin\frac{\alpha}{2}$$ $$G=\sin\alpha-2\sin\frac{\alpha}{2}$$ $$F=-3\sin\alpha+2\sin\frac{\alpha}{2}$$ $$H=\sin^2\alpha$$ and $\alpha\in(0,\pi/2)$. Note that $K>0,G<0,F<0$ when $\alpha\in(0,\pi/2)$.
we assume (1) $\lambda>\frac{-B-\sqrt{B^2-4AC}}{2A}$ where $A=\frac{1}{2}(\frac{n}{2}-1)F,B=-(\frac{n}{2}K+G)\lambda,C=\frac{n}{4}K-\frac{1}{2}K$ holds, in order to guarantee that $E<0$. (2) $\lambda$ is a decreasing function of $n$ such that its limitation is $\frac{-B-\sqrt{B^2-4AC}}{2A}$ when $n\rightarrow\infty$.
Question
I am trying to find the condition on $\lambda$ such that $$\frac{E^2}{P}>\log(\frac{n(n-1)}{2})$$
Where I got stuck at
To analyze the growth of fraction $$\frac{E^2}{P}$$ I observe in the denominator and nominator, there are some terms that we can tell which ones are growing, which ones are decreasing, when $n$ increases. However, there are also terms we cannot tell or they are constant.
Explicitly, I color the growing terms as red and decreasing terms as blue, and then rest (black) ones are the ones we cannot tell or they are constant.
$$\frac{(\frac{F}{4}n\lambda^2\color{blue}{-\frac{F}{2}\lambda^2}-\frac{K}{2}n\lambda\color{blue}{-G\lambda}+\color{red}{\frac{n}{4}K}-\frac{1}{2}K)^2}{-\frac{1}{8}Hn\lambda^3+\color{blue}{\frac{1}{4}H\lambda^3}-\color{red}{4H\lambda^2}-\frac{1}{16}Hn\lambda^4+\color{blue}{\frac{1}{8}H\lambda^4}+\color{red}{\frac{3}{16}Hn}-\frac{3}{8}H}$$
actually, we can throw away some terms and arrive
$$\frac{(\frac{F}{4}n\lambda^2\color{blue}{-\frac{F}{2}\lambda^2}-\frac{k}{2}n\lambda+\color{red}{\frac{n}{4}K})^2}{-\frac{1}{8}Hn\lambda^3-\frac{1}{16}Hn\lambda^4+\color{blue}{\frac{1}{8}H\lambda^4}+\color{red}{\frac{3}{16}Hn}}$$
Thus, in this case, I don't know how to do the analysis. Could someone help me out? Many thanks for your help.
Further information that might needed:
$\frac{-B-\sqrt{B^2-4AC}}{2A}$ is asymptotically equivalent to $\frac{K-\sqrt{K^2-FK}}{F}$ for sufficiently large $n$. Moreover, $\frac{K-\sqrt{K^2-FK}}{F}$ is a function of $\alpha$, and its value is around $0.42$ when $\alpha\in (0,\pi/2)$.