Another proof of Bertrand-Chebyshev Theorem

We prove that the following inequality holds for $n≧2$.
$$n< p_m<2n\ ...(1)$$
Let $p_m$ be the largest prime number among the primes smaller than or equal to $n$. We assume that the following inequality holds.
$$p_m≦n<2n< p_{m+1}$$
$$n< p_{m+1}-p_m$$
According to Cramér's conjecture $p_{m+1}-p_m<(\log p_m)^2$ holds for $m≧5$, $n<(\log p_m)^2$ must hold when the above inequality holds. Since $p_m≦n$ holds,
$$n<(\log n)^2$$
holds. Let $f(n)=(\log n)^2-n$. $f(n)<0$ holds for $n≧11$. The primes between $n$ and $2n$ for $2≦n≦10$ 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 $n≧2$. (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 $m≧6$, 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 $m≧6$. The primes between $n$ and $2n$ for $2≦n≦6$ 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 $n≧2$. (Q.E.D.)