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Let $text{lcm}(x)$ be the least common multiple of ${1,2,3,dots,x}$.

Denis Hanson showed that $text{lcm}(x) < 3^x$ and M. Nair showed that $text{lcm}(x) > 2^x$.

Neither used Bertrand's Postulate in their result. Hanson later showed that there is always a prime between $3n$ and $4n$ without using Nair's result.

The argument for Bertrand's Postulate depends on the following idea:

$$text{lcm}(sqrt{2x})frac{2x#}{x#}ge frac{text{lcm}(2x)}{text{lcm}(x)}$$

where $2x#$ and $x#$ are the primorial for $2x$ and for $x$.

Here is the argument:

• If a prime $sqrt{2x}<ple x$, then it cancels out in $dfrac{text{lcm}(2x)}{text{lcm}(x)}$.




• If a prime $x < p le 2x$, then it divides $dfrac{2x#}{x#}$.




• If $a ge 2$ and $x < p^a le 2x$, then $p^{a-1} < dfrac{x}{p} < x$ and it divides out and $p^{a+1} > 2x$.




• The lemma follows.

Here is the argument for Bertrand's Postulate:

(1) From Hanson and Nair:

$$frac{text{lcm}(2x)}{text{lcm}(x)} > frac{2^{2x}}{3^{x}} = left(frac{4}{3}right)^x$$

(2) Assume that there is no prime between $2x$ and $x$.

(3) Then we have the following:

$$3^{sqrt{2x}} > text{lcm}(sqrt{2x})frac{2x#}{x#} > left(frac{4}{3}right)^x$$

(4) Which simplifies to:

$$frac{ln(4)}{ln(3)} < frac{x+sqrt{2x}}{x}$$

which is invalid for $x ge 30$ since:

$$frac{ln(4)}{ln(3)} > 1.26 > frac{30+sqrt{60}}{30} approx 1.258$$

and

$$frac{d}{dx}left(frac{x+sqrt{2x}}{x}right) = -frac{1}{sqrt{2}x^{3/2}}$$

Am I wrong?

I think that you are right.

I think that the inequality of the idea is true, but it seems that your proof has an error. The third bullet point looks wrong since $xlt p^ale 2x$ does not imply $p^{a-1}lt frac xplt x$.

You've already dealt with the primes $p$ such that $sqrt{2x}lt ple 2x$ in the first and the second bullet points.

Let us consider the case where $$p^ale sqrt{2x}lt p^{a+1}qquadtext{and}qquad p^ble xlt p^{b+1}$$
Here, $p$ is a prime and $a,b$ are positive integers such that $ale b$.

If $a=b$. Then, we get
$$frac{p^{2a}}{2}le xlt p^{b+1}=p^{a+1}implies p^{a-1}lt 2implies a=b=1$$
So, we have
$$ple sqrt{2x}lt p^{2},qquad ple xlt p^{2},qquad p^2le 2xlt 2p^2le p^3$$
So, in this case, the inequality holds.

If $bge a+1$, then we get
$$p^{2a}le 2xlt p^{2a+2}$$
and
$$p^ap^bge p^{2a+1}ge p^{2a}$$
So, in this case, the inequality holds.