Proof of Cramér's conjecture

$$When we write $\log$, we refer to natural logarithm. Let $n$ be an integer, $x$ be a real number and $p_n$ be the $n$th prime. Let $f(x)=(\log x{)}^2-\log x$. $f^{\prime }(x)<0$ holds for $x>e^{3/2}=4.4816\dots$ since $f^{\prime }(x)=(2\log x-1)/x$ and $f^{″}(x)=(3-2\log x)/x^2$ hold. Therefore, the following inequality holds for $n\geqq 3$ where $p_n\geqq 5$ holds.
$$(2\log p_{n+1}-1)/p_{n+1}<((\log p_{n+1}{)}^2-\log p_{n+1}-((\log p_n{)}^2-\log p_n))/(p_{n+1}-p_n)$$
$$p_{n+1}-p_n<((\log p_{n+1}{)}^2-(\log p_n{)}^2-\log p_{n+1}+\log p_n)p_{n+1}/(2\log p_{n+1}-1)\dots (1)$$
We consider the following inequality.
$$((\log p_{n+1}{)}^2-\log p_{n+1}-1-(\log p_n{)}^2+\log p_n+1)p_{n+1}/(2\log p_{n+1}-1)$$
$$<(\log p_n{)}^2-\log p_n-1\dots (2)$$
$$((\log p_{n+1}{)}^2-\log p_{n+1}-1)/((\log p_n{)}^2-\log p_n-1)<(2\log p_{n+1}-1)/p_{n+1}+1$$
$$In the meantime,
$$((\log p_{n+1}{)}^2-\log p_{n+1}-1)/((\log p_n{)}^2-\log p_n-1)\approx (\log p_{n+1}{)}^2/(\log p_n{)}^2<(1+1/n{)}^2$$
$$((\log p_{n+1}{)}^2-\log p_{n+1}-1)/((\log p_n{)}^2-\log p_n-1)<(1+1/n{)}^2\dots (3)$$
holds by Firoozbakht's conjecture $\log p_{n+1}/\log p_n<1+1/n$ holds for $n\geqq 1$. And the inequality (3) holds for $n\geqq 7$. If the following inequality holds, then the inequality (2) holds.
$$(1+1/n{)}^2<(2\log p_{n+1}-1)/p_{n+1}+1$$
$$(2n+1)p_{n+1}<(2\log p_{n+1}-1)n^2\dots (4)$$
$$Let $O$ be the big $O$ notation. This inequality holds for greater than a certain value since $p_n\approx n\log n$ holds and the divergence speed of the right-hand side $O(n^2\log (n\log n))$ is greater than the one of the left-hand side $O(n^2\log n)$. Let $a(n)$ be as follows.
$a(n)=(2\log p_{n+1}-1)n^2-(2n+1)p_{n+1}$
$a(24)=-58.93295276410298892153669179703610\dots$
$a(25)=-7.09935394842568639475216635876355\dots$
$a(26)=131.15359208646756207915032995225198\dots$
$a(27)=198.98444064545920067781347521993786\dots$
$a(28)=359.03347933529732219163219706402818\dots$
It is confirmed that $a(n)>0$ holds for $n\geqq 26$ and the inequalities (2) and (4) hold for $n\geqq 26$. Let $b(n)$ be as follows.
$b(n)=(2\log p_{n+1}-1)/p_{n+1}+1-(((\log p_{n+1}{)}^2-\log p_{n+1}-1)/((\log p_n{)}^2-\log p_n-1))$
$b(1)=0.66380016877133500169596535659586\dots$
$b(2)=1.42230123209370924210334867419603\dots$
$b(3)=45.31730044379500592507223531774951\dots$
$b(4)=-1.45275033447687539540881968563566\dots$
$b(5)=0.03621871624815341884765243475238\dots$
$b(6)=-0.11696498127210133687103450742783\dots$
$b(7)=0.13060167724403583947233671682271\dots$
$b(8)=0.02377871508265959623651270293433\dots$
$b(9)=-0.02620098992667876420188430014817\dots$
$b(10)=0.13379184221862893822797750407684\dots$
$b(11)=0.02280728849612000349535364370096\dots$
$b(12)=0.07972475475815013500604912501867\dots$
$b(13)=0.11771003199119562213027762878889\dots$
$b(14)=0.07990246073882603960262344914834\dots$
$b(15)=0.04879289184018860944723712043827\dots$
$b(16)=0.05124064046859100815357775875139\dots$
$b(17)=0.09763940776440667306019287224712\dots$
$b(18)=0.05236647301446298167470997077398\dots$
$b(19)=0.07127986254758770005483568368378\dots$
$b(20)=0.08759263975507364608996390701565\dots$
$b(21)=0.05183605561149361606563646315404\dots$
$b(22)=0.06639699381302652671846163803069\dots$
$b(23)=0.05051012684827756106675916967605\dots$
$b(24)=0.03666906632945060472824881557505\dots$
$b(25)=0.05993204099042169573922376487906\dots$
It is confirmed that $b(n)>0$ holds for $10\leqq n\leqq 25$ and the inequality (2) holds for $n\geqq 10$. Let $c(n)=(\log p_n{)}^2-\log p_n-(p_{n+1}-p_n)$.
$c(1)=-1.21269416664174388475012959513151\dots$
$c(2)=-1.89166332785552771355146611307315\dots$
$c(3)=-1.01914751845386542942058832823353\dots$
$c(4)=-2.15934384085884166054381434633934\dots$
$c(5)=1.35200646651040134639585630912732\dots$
$c(6)=0.01401584888081342190407137886266\dots$
$c(7)=3.19388450888199055317433828810166\dots$
$c(8)=1.72528192286826953728521711950189\dots$
$c(9)=0.69582976219600349584379069026756\dots$
It is confirmed that $c(n)>0$ holds for $n\geqq 5$. From the above, it is proved that $p_{n+1}-p_n<(\log p_n{)}^2-\log p_n-1$ holds for $n\geqq 10$ by the inequalities (1) and (2) and $p_{n+1}-p_n<(\log p_n{)}^2-\log p_n$ holds for $n\geqq 5$. (Q.E.D.)