I came across the following asymptotic problem in my reseach, however I don't know how to answer it.
Let $C_0,C_1,C_2,C_3$ be absolute constant (do not depend on $n$), and $p(n)$ is a function of $n$.
We need to find $p(n)$ such that the following two conditions hold:
(1) $p(n)$ that is larger than the following function, i.e., $$\tiny p(n)>\frac{-(4+(n-2)(2C_1+C_2))+\sqrt{((4+(n-2)(2C_1+C_2)))^2-4\cdot (-(n-2)(C_1+C_2-C_3))\cdot(-2C_0-(n-2)C_1)}}{2(-(n-2)(C_1+C_2-C_3))}$$ It is okay if this only holds for sufficiently large $n$.
(2) Let $$\tiny E(p(n),n):=(-2C_0-(n-2)C_1+(4+(n-2)(2C_1+C_2))\cdot p-(n-2)(C_1+C_2-C_3)p^2)^2$$ and $$\tiny F(p(n),n):= 4C_0^2\cdot (-4p^2+4p)+C_1^2\cdot \frac{1}{8}(n-2)4(-p+1)p+(n-1)(1-(1-2p)^4)+C_2\cdot \frac{1}{4}(n-1)\big(1-(1-2p)^4\big)+C_3^2 \frac{1}{8}(n-2)4(-p+1)p+(n-1)(1-(1-2p)^4)$$
we need $$E\succ F$$ for sufficiently large $n$. In other words, $$\lim_{n\rightarrow+\infty}\frac{E}{F}=+\infty$$
What I tried
Should I assume the following and then plug it into $\frac{E}{F}$?
$$\tiny p(n)=\frac{-(4+(n-2)(2C_1+C_2))+\sqrt{((4+(n-2)(2C_1+C_2)))^2-4\cdot (-(n-2)(C_1+C_2-C_3))\cdot(-2C_0-(n-2)C_1)}}{2(-(n-2)(C_1+C_2-C_3))}+g(n)$$
With the help of a CAS, we find
$$ \frac{F}{E}\sim\frac{1}{n} \frac{N}{D} \qquad, \qquad n \to \infty $$
Where
$$ N= \left(-4 C_2-32\right) p^4+\left(8 C_2+64\right) p^3+\left(-\frac{C_1^2}{2}-6 C_2-\frac{C_3^2}{2}-48\right) p^2+\left(\frac{C_1^2}{2}+\frac{C_3^2}{2}+2 C_2+16\right) p \\ D=\left(C_1+C_2-C_3\right){}^2 p^4-2 \left(2 C_1+C_2\right) \left(C_1+C_2-C_3\right) p^3+\left(\left(2 C_1+C_2\right){}^2+2 C_1 \left(C_1+C_2-C_3\right)\right) p^2-2 C_1 \left(2 C_1+C_2\right) p+C_1^2 $$
Consider cases: If $p\to \infty$ then the $4$ powers of $p$ in the numerator match the $4$ powers in the denominator, so $F/E\sim 1/n \to 0$. If $p\to 0$ then the numerator vanishes and the denominator stays finite, so $F/E\to 0$. If $p\to \text{const}$ then $F/E\sim1/n\to 0$. So for any choice of $p$, your second condition is fulfilled- as long as the appropriate leading term doesn't vanish due to an unfortunate relation between the $C_i$ (such as $C_1+C_2-C_3=0$).
For your first condition, let $Q= [\text{RHS expression}]$. We have
$$ Q\sim K=\frac{2C_1+C_2-\sqrt{C_2^2+4C_1 C_3}}{2(C_1+C_2-C_3)} \qquad, \qquad n \to \infty $$
So we want to choose $p=K$. However, $Q$ may approach $K$ from above or below, and the inequality $p>Q$ means we must ensure that the difference $p-Q>0$. The difference $K-Q$ shows up at order $1/n$. If the coefficient at this order is positive, we can choose $p=K$. If the coefficient is negative, we can add a small positive term to $K$. The smallest term we can add that fixes the difference will be just a little larger than $1/n$. So the 'best' choice, written compactly to take care of either case above, is
$$ p(n)=K+n^{\epsilon-1}H(X) $$
Where $0<\epsilon\ll 1$, $H$ is the Heaviside step function, and $-X$ has the sign of the difference $K-Q$, specifically:
$$ X=\left( C_1+C_2-C_3\right)\left(\sqrt{C_2^2+4 C_1 C_3}+\left(C_0-2\right) C_1+\left(C_0-1\right) C_2-C_0 C_3\right) $$