Proof of Cramér's conjecture

$ $When we write $\log$, we refer to natural logarithm. Let $n$ be an integer, $x$ be a real number and $p_n$ be the $n$th prime. We will prove that the following inequality holds for $n≧10$.
$$p_{n+1}-p_n<(\log\ p_n)^2-\log\ p_n-1\ ...(1)$$
Let $f(x)=(\log\ x)^2-\log\ x-1$. $f'(x)>0$ holds for $x>e^{1/2}=1.6487…$ since $f'(x)=(2 \log\ x-1)/x$ holds. Therefore, the following inequality holds for $n≧1$ when $p_n≧2$ holds.
$$(2\log\ p_n-1)/p_n<((\log\ p_{n+1})^2-\log\ p_{n+1}-1-((\log\ p_n)^2-\log\ p_n-1))/(p_{n+1}-p_n)$$
$$p_{n+1}-p_n<((\log\ p_{n+1})^2-\log\ p_{n+1}-1-((\log\ p_n)^2-\log\ p_n-1))p_n/(2\log\ p_n-1)\ ...(2)$$
We consider the following inequality.
$$((\log\ p_{n+1})^2-\log\ p_{n+1}-1-((\log\ p_n)^2-\log\ p_n-1))p_n/(2\log\ p_n-1)<(\log\ p_n)^2-\log\ p_n-1$$
$$((\log\ p_{n+1})^2-\log\ p_{n+1}-1)/((\log\ p_n)^2-\log\ p_n-1)<(2\log\ p_n-1)/p_n+1\ ...(3)$$
$ $In the meantime, $((\log\ p_{n+1})^2-\log\ p_{n+1}-1)/((\log\ p_n)^2-\log\ p_n-1)≈(\log\ p_{n+1}/\log\ p_n)^2$ holds. Therefore, by Firoozbakht's conjecture $\log\ p_{n+1}/\log\ p_n<1+1/n$ holds for $n≧1$, the following inequality holds for values greater than a certain value.
$$((\log\ p_{n+1})^2-\log\ p_{n+1}-1)/((\log\ p_n)^2-\log\ p_n-1)<(1+1/n)^2\ ...(4)$$
Let $a(n)$ be as follows.
$a(n)=(1+1/n)^2-((\log\ p_{n+1})^2-\log\ p_{n+1}-1)/((\log\ p_n)^2-\log\ p_n-1)$
$a(5)=0.15853419971560930560865436681925...$
$a(6)=-0.03034955769701564696575099841944...$
$a(7)=0.17941475999558131199555402259606...$
$a(8)=0.06023030500186397094896897842909...$
$a(9)=0.01062237130882196274418208161315...$
It is confirmed that the inequality (4) holds for $n≧7$. If the following inequality holds, then the inequality (3) holds.
$$(1+1/n)^2<(2\log\ p_n-1)/p_n+1\ ...(5)$$
$$2/n+1/n^2<(2\log\ p_n-1)/p_n$$
$ $The function $f'(x)$ is decreasing monotonically for $x>e^{3/2}=4.4816...$ since $f''(x)=(3-2\log\ x)/x^2$ holds. Because $p_n< n(\log\ n+\log(\log\ n)-1/2)$ holds for $n≧20$,
$$(2\log(n(\log\ n+\log(\log\ n)-1/2))-1)/(n(\log\ n+\log(\log\ n)-1/2))<(2\log\ p_n-1)/p_n$$
holds for $n≧20$. We consider the case where the following inequality holds.
$$2/n+1/n^2<(2\log(n(\log\ n+\log(\log\ n)-1/2))-1)/(n(\log\ n+\log(\log\ n)-1/2))$$
$$(2n+1)n(\log\ n+\log(\log\ n)-1/2)<(2\log(n(\log\ n+\log(\log\ n)-1/2))-1)n^2$$
$$(2n^2+n)(\log\ n+\log(\log\ n))-n^2-n/2<2(\log n+\log(\log\ n+\log(\log\ n)-1/2))n^2-n^2$$
$$2\log(\log\ n)n^2+(\log\ n+\log(\log\ n))n<2(\log(\log\ n+\log(\log\ n)-1/2))n^2+n/2$$
Let $O$ be a big O notation. This inequality holds for values greater than a certain value since the divergence speed of the right-hand side $O(n^2\log(\log(n\log\ n)))$ is faster than the one of the left-hand side $O(n^2\log(\log\ n))$. Let $b(n)$ be as follows.
$b(n)=2(\log(\log\ n+\log(\log\ n)-1/2))n^2+n/2-(2\log(\log\ n)n^2+(\log\ n+\log(\log\ n))n)$
$b(8)=-4.94800797366010654626386295300024...$
$b(9)=-2.45903905297587289380618226147847...$
$b(10)=0.72670759996052204074970382206822...$
$b(11)=4.62973188882811433737389178369106...$
$b(12)=9.26440469623080983321949580015932...$
It is confirmed that $b(n)>0$ holds for $n≧10$ and the inequality (5) holds for $n≧20$. Hence, when $n≧20$ holds, the inequality (3) holds by the inequalities (4) and (5) and the inequality (1) holds by the inequalities (2) and (3). Let $c(n)$ be as follows.
$c(n)=(\log\ p_n)^2-\log\ p_n-1-(p_{n+1}-p_n)$
$c(1)=-2.21269416664174388475012959513151...$
$c(2)=-2.89166332785552771355146611307315...$
$c(3)=-2.01914751845386542942058832823353...$
$c(4)=-3.15934384085884166054381434633934...$
$c(5)=0.35200646651040134639585630912732...$
$c(6)=-0.98598415111918657809592862113733...$
$c(7)=2.19388450888199055317433828810166...$
$c(8)=0.72528192286826953728521711950189...$
$c(9)=-0.30417023780399650415620930973243...$
$c(10)=4.97138537665782296909198804023450...$
$c(11)=1.35828091608256337148497731718031...$
$c(12)=4.42781025921069847108093782340415...$
$c(13)=7.07704542790219612524012460543752...$
$c(14)=5.38542619459970493585965424305425...$
$c(15)=3.97348895324365734391949940247134...$
$c(16)=4.79292596526524744134939557344027...$
$c(17)=9.54877416254746874468494698073696...$
$c(18)=5.78841006296990061293722388345165...$
$c(19)=8.47474740416989731223131306006849...$
It is confirmed that $c(n)>0$ holds for $n≧10$ and $c(n)>-1$ holds for $n≧5$. 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.)