Another proof of Cramér's conjecture

$ $When we write $\log$, we refer to natural logarithm. Let $e$ be Napier's constant, $n$ be an integer, $x$ and $y$ be real numbers and $p_n$ be the $n$th prime. We prove that the following inequality holds for $n≧10$.
$$p_{n+1}-p_n<(\log\ p_n)^2-\log\ p_n-1\ ...(1)$$
$$p_{n+1}/p_n<1+((\log\ p_n)^2-\log\ p_n-1)/p_n$$
$$\log\ p_{n+1}-\log\ p_n<\log(1+((\log\ p_n)^2-\log\ p_n-1)/p_n)$$
Since $y/(x+y)<\log(x+y)-\log\ x$ holds for $x>0$ and $y>0$, in order for the inequality (1) to hold, setting $x=p_n$ and $y=(\log\ p_n)^2-\log\ p_n-1$, the following inequality must hold.
$$\log\ p_{n+1}-\log\ p_n<((\log\ p_n)^2-\log\ p_n-1)/((\log\ p_n)^2-\log\ p_n-1+p_n)$$
$$\log\ p_{n+1}/\log\ p_n<1+(\log\ p_n-1-1/\log\ p_n)/((\log\ p_n)^2-\log\ p_n-1+p_n)$$
$ $According to the Firoozbakht's conjecture $\log\ p_{n+1}/\log\ p_n<1+1/n$ holds for $n≧1$, the following inequality must hold.
$$1/n<(\log\ p_n-1-1/\log\ p_n)/((\log\ p_n)^2-\log\ p_n-1+p_n)$$
$$\log\ p_n+p_n\log\ p_n/((\log\ p_n)^2-\log\ p_n-1)< n\ ...(2)$$
Let $f_1(x)=\log\ x+x\log\ x/((\log\ x)^2-\log\ x-1)$. $f_1'(x)=1/x+((\log\ x)^3-2(\log\ x)^2-\log\ x-1)/((\log\ x)^2-\log\ x-1)^2$ holds. Let $f_2(x)=x^3-2x^2-x-1$. $f_2(x)>0$ holds for $x≧3$ since $f_2'(x)=3x^2-4x-1$ holds, $f_2'(x)>0$ holds for $x=(2+ \sqrt{7})/3=1.5485...$ and $f_2(3)=5$ holds. Therefore, $f_1'(x)>0$ holds for $x≧e^3=20.0855...$ and the left-hand side of the inequality (2) is monotonically increasing when $n≧9$ holds. For the range of $n$ where the inequality obtained by replacing $p_n$ in the inequality (2) with an upper bound of $p_n$ holds, the inequality (2) holds in the range.
$ $By the inequality (2),
$$p_n<(n-\log\ p_n)(\log\ p_n-1-1/\log\ p_n)$$
holds. Let $O$ be big O notation. This inequality holds for values greater than a certain value since the divergence rate of the right-hand side of the inequality $O(n\log(n\log(n\log\ n)))$ is greater than the one of the left-hand side $O(n\log(n\log\ n))$. According to Dusart’s inequality $p_n< n(\log\ n+\log(\log\ n)−0.9484)$ holds for $n≧39017$, let $f_3(x)$ be as follwing.
$f_3(n)=n-(\log(n(\log\ n+\log(\log\ n)−0.9484))+(n(\log\ n+\log(\log\ n)−0.9484))$
$/(\log(n(\log\ n+\log(\log\ n)−0.9484))-1-1/\log(n(\log\ n+\log(\log\ n)−0.9484))))$
It is confirmed that $f_3(n)>0$ holds for $n≧222633$ by numerical computation. Let $f_4(n)$ be as follows.
$f_4(n)=(n-\log\ p_n)(\log\ p_n-1-1/\log\ p_n)-p_n$
It is confirmed that $f_4(n)>0$ holds for $85036≦n≦222632$. Let $f_5(n)$ be as follows.
$f_5(n)=(\log\ p_n)^2-\log\ p_n-1-(p_{n+1}-p_n)$
Numerical computation shows that $f_5(n)>0$ holds for $10≦n≦85035$ and $f_5(n)>-1$ holds for $5≦n≦9$. From the above, it is proved that the inequality (1) holds for $n≧10$ and $p_{n+1}-p_n<(\log\ p_n)^2-\log\ p_n$ holds for $n≧5$. (Q.E.D.)