The following is an MO-adopted extract from my lectures.

Euclid's theorem about the infiniteness of primes
can be written as
$$
\pi(x)\to+\infty \qquad\text{as}\quad x\to+\infty.
$$
However, this theorem tells us nothing about how fast the function
$\pi(x)$ increases with $x$.
In 1808 Legendre published a discovered empirical formula, namely,
$$
\pi(x)\approx\frac x{\ln x-1.08366};
$$
some years later Gauss noted that $x/(\ln x-1)$ and even
$$
\int_2^x\frac{dt}{\ln t}
$$
is a much better approximation to $\pi(x)$ for larger $x$.
In 1850 Chebyshev published his work
containing the following result.

**Theorem** (Chebyshev).
There exist two constants $A$ and $B$,
$0< A< 1< B$, such that for all $x\ge2$ the following bounds are valid:
$$
\frac{Ax}{\ln x}< \pi(x)< \frac{Bx}{\ln x}.
\qquad(1)
$$

Chebyshev actually proved the theorem by showing that one can take
$A=0.921$ and $B=1.106$ for $x\ge x_0$, so that
$A$ and $B$ are very close to $1$. In our simplified proof
we will get more modest estimates for $A$ and $B$.
Note however that for the proof of the prime number theorem
one needs the existence result for
these constants. Chebyshev's theorem states the correct
order of growth of the function $\pi(x)$
as $x\to+\infty$.

We need some properties of the function
$$
T(x)=\sum_p\ln p\biggl(\biggl[\frac xp\biggr]
+\biggl[\frac x{p^2}\biggr]
+\biggl[\frac x{p^3}\biggr]+\dots\biggr),
$$
where $[\cdot]$ denotes the integer part of a real number.

**Lemma 1.**
For positive integers $n$ we have the equality
$T(n)=\ln n!$.

*Proof.*
Factorise $n!$ into product of primes,
$n!=\prod_pp^{\nu_p}$, and use
the formula
$$
\operatorname{ord}_p n!=\biggl[\frac np\biggr]
+\biggl[\frac n{p^2}\biggr]
+\biggl[\frac n{p^3}\biggr]+\dots
$$
(the sum is in fact finite!) to see that
$$
\nu_p=\biggl[\frac np\biggr]
+\biggl[\frac n{p^2}\biggr]
+\biggl[\frac n{p^3}\biggr]+\dots
\qquad\text{for each prime $p$}.
$$
Taking the logarithm of the factorisation of $n!$
we deduce the required equality.

**Lemma 2.**
The function $T(n)$ (of natural argument $n$) is nondecreasing.

*Proof.*
It follows from Lemma 1 that
$$
T(n+1)-T(n)=\ln(n+1)!-\ln n!=\ln(n+1)> 0.
$$

**Lemma 3.**
If $[x]=n$, then $T(x)=T(n)$.

*Proof.*
Let $n=[x]$, $\alpha=\{x\}$ (the fractional part of $x$, hence $0< \alpha< 1$),
and let $m$ be a power of prime $p$.
If $[n/m]=q$, then $n=mq+r$ where $0\le r\le m-1$.
Therefore, $x=n+\alpha=mq+(r+\alpha)$ where
$0\le r+\alpha< m$; consequently,
$$
\biggl[\frac xm\biggr]
=\biggl[q+\frac{r+\alpha}m\biggr]
=q+\biggl[\frac{r+\alpha}m\biggr]
=q=\biggl[\frac nm\biggr].
$$
From this inequality and from the definition of $T(x)$
the required property follows immediately.

We will also need the following elementary inequality:
for $n\ge3$,
$$
(n+1)^3\le2^{2n}.
$$

**Lemma 4.**
For $x\ge6$ the following estimates take place:
$$
\frac{\ln2}2\cdot x< T(x)-2T\biggl(\frac x2\biggr)
< \frac{4\ln2}3\cdot x.
$$

*Proof.*
Denote $n=[x/2]\ge3$, so that $2n\le x< 2n+2$. Then by Lemma 3,
$$
T(x)-2T\biggl(\frac x2\biggr)
=T(2n)-2T(n) \quad \text{if $2n\le x< 2n+1$}, \qquad\text{and}\qquad
=T(2n+1)-2T(n) \quad \text{if $2n+1\le x< 2n+2$},
$$
hence with the help of Lemma 2 we obtain
$$
T(2n)-2T(n)
\le T(x)-2T\biggl(\frac x2\biggr)
\le T(2n+1)-2T(n).
\qquad(2)
$$

By Lemma 1 for $n\ge3$ we find that
$$
T(2n)-2T(n)
=\sum_{k=1}^{2n}\ln k-2\sum_{k=1}^n\ln k
=\sum_{k=n+1}^{2n}\ln k-\sum_{k=1}^n\ln k
$$
$$
=\ln\frac{n+1}{1}+\ln\frac{n+2}{2}+\dots+\ln\frac{n+n}{n}
\ge(n+1)\ln2
\gt \frac{x}{2}\ln 2,
\qquad (3)
$$
so that
$$
\ln\frac{n+1}1\ge\ln4=2\ln2,
\qquad
\ln\frac{n+k}k=\ln\biggl(1+\frac nk\biggr)\ge\ln2
\quad\text{for } k=2,3,\dots,n,
$$
and $x/2< n+1$.

On the other hand, for $n\ge3$,
$$
\frac{(2n+1)!}{n!^2}
=(n+1)\cdot\frac{(2n+1)!}{n!(n+1)!}
=\frac{n+1}2\cdot\biggl(\binom{2n+1}n+\binom{2n+1}{n+1}\biggr)
$$
$$
< \frac{n+1}2\sum_{k=0}^{2n+1}\binom{2n}k
=\frac{n+1}2\cdot(1+1)^{2n+1}=2^{2n}(n+1)
$$
$$
\le2^{2n+2n/3}=2^{8n/3}
$$
(on the last step we use the above elementary inequality).
Thus, by Lemma 1
$$
T(2n+1)-2T(n)=\ln\frac{(2n+1)!}{n!^2}
< \frac{8n}3\ln2\le\frac{4x}3\ln2.
\qquad(4)
$$
Substituting the inequalities (3) and (4)
into (2), we arrive at the desired claim.

**Lemma 5.**
For a real number $\alpha$, the difference
$[2\alpha]-2[\alpha]$ is either $0$ or $1$.

The proof of the lemma is straightforward.

*Proof of Chebyshev's theorem.
Lower bound.*
Using the definition of $T(x)$ we obtain
$$
T(x)-2T\biggl(\frac x2\biggr)
=\sum_{p\le x}\ln p\biggl(
\biggl(\biggl[\frac xp\biggr]-2\biggl[\frac x{2p}\biggr]\biggr)
+\biggl(\biggl[\frac x{p^2}\biggr]-2\biggl[\frac x{2p^2}\biggr]\biggr)
+\dots\biggr),
\qquad(5)
$$
since all the terms on the right-hand side vanish for $p> x$.
The groups
$$
\biggl(\biggl[\frac x{p^k}\biggr]-2\biggl[\frac x{2p^k}\biggr]\biggr)
$$
also vanish for $p^k> x$, whence the coefficient of $\ln p$
on the right-hand side of (5) can be written as
$$
\sum_{k=1}^s
\biggl(\biggl[\frac x{p^k}\biggr]-2\biggl[\frac x{2p^k}\biggr]\biggr),
$$
for $p^s\le x$, that is, for $s\le\dfrac{\ln x}{\ln p}$.
By Lemma 5 each term in parenthesis on the right-hand side of (5)
is at most $1$, and the number of such terms is at most $s$,
hence
$$
T(x)-2T\biggl(\frac x2\biggr)
\le\sum_{p\le x}\ln p\cdot\frac{\ln x}{\ln p}
=\ln x\cdot\sum_{p\le x}1
=\pi(x)\ln x.
$$
Consequently, by Lemma 4
$$
\pi(x)\ln x> \frac{\ln2}2x,
$$
and the desired lower estimate is deduced with $A=\frac12\ln2$.

*Upper bound.*
Since all the terms on the right-hand side of (5)
are non-negative, leaving those for which $p> x/2$ does not
increase the whole sum. For $x/2< p\le x$ we have
$$
\biggl(\biggl[\frac xp\biggr]-2\biggl[\frac x{2p}\biggr]\biggr)
+\biggl(\biggl[\frac x{p^2}\biggr]-2\biggl[\frac x{2p^2}\biggr]\biggr)
+\dots
=1,
\qquad(6)
$$
because for such $p$ and $x\ge6$ there holds $2p> x$ and $p^2> x^2/4> x$,
that is, the left-hand side of (6) involves a unique nonzero term,
$[x/p]$. Thus, from (5) we obtain
$$
T(x)-2T\biggl(\frac x2\biggr)
\ge\sum_{x/2< p\le x}\ln p
\gt \ln\frac x2\cdot\sum_{x/2< p\le x}1
=\biggl(\pi(x)-\pi\biggl(\frac x2\biggr)\biggr)\ln\frac x2
$$
$$
=\pi(x)\ln x-\pi\biggl(\frac x2\biggr)\ln\frac x2-\pi(x)\ln2
$$
$$
\ge\pi(x)\ln x-\pi\biggl(\frac x2\biggr)\ln\frac x2-x\ln2
$$
(where we applied the trivial bound $\pi(x)\le x$),
whence
$$
\pi(x)\ln x-\pi\biggl(\frac x2\biggr)\ln\frac x2< T(x)
-2T\biggl(\frac x2\biggr)+x\ln2< \biggl(\frac{4\ln2}3+\ln2\biggr)x< 2x
\qquad(7)
$$
for $x\ge6$. For any $x> 0$ there exists a positive integer $s$
such that $x/2^s< 2$, so that $\pi(x/2^s)=0$.
Successively applying the inequalities (7) we find that
$$
\pi(x)\ln x
< 2x+\pi\biggl(\frac x2\biggr)\ln\frac x2
< 2x+2\cdot\frac x2+\pi\biggl(\frac x4\biggr)\ln\frac x4
< \dots
$$
$$
< 2x+2\cdot\frac x2+\dots+2\cdot\frac x{2^{s-1}}
+\pi\biggl(\frac x{2^s}\biggr)\ln\frac x{2^s}
$$
$$
< 2x\biggl(1+\frac12+\frac14+\dots\biggr)
=4x,
$$
which gives us the required estimate with $B=4$.

*Remarks.*
In order to get $A$ and $B$ closer to $1$,
Chebyshev considered the expression
$$
T(x)-T\biggl(\frac x2\biggr)
-T\biggl(\frac x3\biggr)
-T\biggl(\frac x5\biggr)
+T\biggl(\frac x{30}\biggr)
$$
instead of $T(x)-2T(x/2)$; in addition, he first
deduced estimates for the functions
$$
\vartheta(x)=\sum_{p\le x}\ln p
\qquad\text{and}\qquad
\psi(x)=\sum_{p\le x}\biggl[\frac{\ln x}{\ln p}\biggr]\ln p,
$$
and then translated them for $\pi(x)$.
The functions $\vartheta(x)$ and $\psi(x)$ are now known as
*Chebyshev's functions*.