Another proof of Bertrand-Chebyshev Theorem

We prove that the following inequality holds for $n\geqq 2$.
$$n<p_m<2n...(1)$$
Let $p_m$ be the largest prime number among the primes smaller than or equal to $n$. We assume that the following inequality holds.
$$p_m\leqq n<2n<p_{m+1}$$
$$n<p_{m+1}-p_m$$
According to Cramér's conjecture $p_{m+1}-p_m<(\log p_m{)}^2$ holds for $m\geqq 5$, $n<(\log p_m{)}^2$ must hold when the above inequality holds. Since $p_m\leqq n$ holds,
$$n<(\log n{)}^2$$
holds. Let $f(n)=(\log n{)}^2-n$. $f(n)<0$ holds for $n\geqq 11$. The primes between $n$ and $2n$ for $2\leqq n\leqq 10$ are as follows.
$2<3<4$
$3<5<6$
$4<5,7<8$
$5<7<10$
$6<7,11<12$
$7<11,13<14$
$8<11,13<16$
$9<11,13,17<18$
$10<11,13,17,19<20$
From the above, it is proved that the inequality (1) holds for $n\geqq 2$. (Q.E.D.)

Second proof

According to Dusart's inequality $m(\log m+\log (\log m)-1)<p_m<m(\log m+\log (\log m))$ holds for $m\geqq 6$, the inequality (1) holds if the following inequalities hold.
$$m(\log m+\log (\log m))<2n$$
$$n<m(\log m+\log (\log m)-1)$$
Let $f(m)$ be as follows.
$f(m)=m(\log m+\log (\log m)-1)-m(\log m+\log (\log m))/2-1$
By the inequalities described above, the inequality (1) holds if $f(m)>0$ holds. The divergence rate of the first term of the above function is greater than the second one, $f(m)>0$ for a certain value.
$f(4)=-1.57414275780365679752148558771627...$
$f(5)=-0.78669273059697251094180863304310...$
$f(6)=0.12487265003214289616583688323301...$
$f(7)=1.14073985871756399965767447940122...$
$f(8)=2.24616363906512517053800977225754...$
It is confirmed that $f(m)>0$ holds for $m\geqq 6$. The primes between $n$ and $2n$ for $2\leqq n\leqq 6$ are described above.
$2<3<4$
$3<5<6$
$4<5,7<8$
$5<7<10$
$6<7,11<12$
From the above, it is proved that the inequality (1) holds for $n\geqq 2$. (Q.E.D.)