How to prove $s > \frac{11}{2}t-3t\ln t$ in this problem The problem is:

Given $f(x)=\frac{\ln x}{x}$, line $l$ is the tangent of curve
$y=f(x)$ at $(t,f(t))$,  and intersects the curve at another point
$(s,f(s))$ where $s<t$.
(1) Find the range of $t$;
(2) (i) Prove $\ln x\le 1 +\frac{1}{e}(x-e)-\frac{1}{2e^2}(x-e)^2+\frac{1}{3e^3}(x-e)^3$;
$\quad$ (ii) Prove $s>\frac{11}{2}t-3t\ln t$.

Now I have solved the problem (1) and (2)(i), but I can't figure out the proof of (2)(ii).
In problem (1) we can get $t\in(e^{3/2},+\infty)$, and the intersection is the null point of the function $F(x)=\frac{\ln x}{x}-\frac{1-\ln t}{t^2}(x-t)-\frac{\ln t}{t}$.
Therefore, we have: $\color{red}{\frac{\ln s}{s}-\frac{1-\ln t}{t^2}(s-t)-\frac{\ln t}{t}=0}$, where $s<t$ and $t>e^{3/2}$. Then we intend to prove $s>\frac{11}{2}t-3t\ln t$.
Maybe the inequality in (2)(i) is helpful for enlarging and reducing the inequality, but I can't find how to use it properly.

For (2)-(ii):
From question (1), we have $t > \mathrm{e}^{3/2}$.
We only need to consider the case when $\frac{11}{2}t - 3t\ln t > 0$, i.e. $t < \mathrm{e}^{11/6}$.
Let
$$G(x) := x F(x) = \ln x - \frac{1-\ln t}{t^2}(x-t)x - \frac{x\ln t}{t}.$$
We have $G(s) = 0$.
We have
$$G'(x) = \frac{(t - x)(t + 2x - 2x\ln t)}{xt^2}.$$
Let $x_0 = \frac{t/2}{\ln t - 1}$.
Since $t > \mathrm{e}^{3/2}$,
we have $x_0 < t$.
We have $G'(x) > 0$ on $(0, x_0)$.
Let $x_1 = \frac{11}{2}t - 3t\ln t$.
Since $t > \mathrm{e}^{3/2}$, we have $x_1 < x_0$.
We can prove that $G(x_1) < 0$ (see the remarks at the end).
Since $G'(x) > 0$ on $(0, x_0)$,
using $G(x_1) < 0$ and $0 < x_1 < x_0$ and $G(s) = 0$,
we have $x_1 < s$.
We are done.

Remarks:
Using $\frac{x_1}{t} = \frac{11}{2} - 3\ln t$, we have
\begin{align*}
 G(x_1) &= \ln x_1 - \frac{1-\ln t}{t^2}(x_1-t)x_1 - \frac{x_1\ln t}{t} \\
 &= \ln t + \ln(11/2 - 3\ln t) - (1 - \ln t)\cdot (11/2 - 3\ln t - 1)(11/2 - 3\ln t)\\
 &\qquad - (11/2 - 3\ln t)\ln t\\
 &= y + \ln(11/2 - 3y) - (1 - y)\cdot (11/2 - 3y - 1)(11/2 - 3y) - (11/2 - 3y)y \\
 &= \ln(11/2 - 3y) + 9y^3 - 36y^2 + \frac{201}{4}y - \frac{99}{4}
\end{align*}
where $y = \ln t \in(3/2, 11/6)$.
Let
$$g(y) := \ln(11/2 - 3y) + 9y^3 - 36y^2 + \frac{201}{4}y - \frac{99}{4}.$$
We have
$$g'(y) = - \frac{81(2y - 3)^3}{4(11 - 6y)}.$$
Thus, $g'(y) < 0$ on $(3/2, 11/6)$.
Also, $g(3/2) = 0$.
Thus, $g(y) < 0$ on $(3/2, 11/6)$.
Thus, $G(x_1) < 0$.
We are done.
