Obvious from the non-negativity of the left-hand side of 1.

Since \(q_i {\lt} n\), the prime \(q_i\) divides \(L_n\) exactly once (as \(q_i^2 {\gt} n\)). Hence we may write \(L_n = q_1 q_2 q_3 L'\) where \(L'\) is the quotient obtained by removing these prime factors. By construction, \(q_i \nmid L'\) for each \(i\).

Use the multiplicativity of \(\sigma (\cdot )\) and the fact that each \(q_i\) occurs to the first power in \(L_n\). Then

Dividing by \(L_n = L' \prod _{i=1}^3 q_i\) gives

This is division with remainder.

The first item is by construction of the division algorithm. For the second, note that

since \(r{\gt}0\). For the third,

Since \(0 {\lt} r {\lt} 4p_1p_2p_3\) and \(4p_1p_2p_3 {\lt} q_1q_2q_3\), we have

so

By Lemma 132,

Hence

Multiplying both sides by \(M\) gives

and hence

since \(M/L_n{\lt}1\) and both sides are integers.

Insert 3 and 2 into the desired inequality and compare with the assumed bound 1; this is a straightforward rearrangement.

If the exponent of \(p_i\) in \(L'\) is \(e\), the contribution of \(p_i\) to \(\sigma (L')/L'\) is

Increasing the exponent to \(e+1\) multiplies this factor by

Since all terms \(1,p_i,\dots \) are positive, replacing \(e\) by \(1\) only decreases this ratio. Thus

In our application, the exponent is exactly increased by \(1\), and we may bound from below by the case \(e=1\), giving the claimed factor

Since \(2^k \le n\), the number \(2^k\) divides \(\lcm (1,2,\dots ,n)\), hence \(2^k\) divides \(L_n\) and thus also divides \(L'\). The definition of \(M\) multiplies \(L'\) by \(4 = 2^2\), so the exponent of \(2\) in \(M\) is at least \(k+2\).

The ratio of contributions of \(2\) to \(\sigma (M)/M\) and \(\sigma (L')/L'\) is

A direct calculation yields

giving the first inequality in 4. Finally, since \(2^k \le n {\lt} 2^{k+1}\), we have \(2^{k+3} {\lt} 8n\), so

giving the second inequality.

Since \(m\) is a positive integer, any extra primes (or higher exponents) it introduces appear as multiplicative factors of the form

Hence they can only increase the value of \(\sigma (M)/M\).

Multiply the contributions from Lemma 135 for \(i=1,2,3\), from Lemma 136 for the prime \(2\), and note that Lemma 137 allows us to ignore any additional (non-decreasing) contribution from \(m\). This gives exactly the stated lower bound.

By Lemma 138, the condition 3 holds. By Lemma 134 this implies

Applying Lemma 133, we obtain \(\sigma (M) \ge \sigma (L_n)\) with \(M {\lt} L_n\), so \(L_n\) cannot be highly abundant.

Apply Theorem 50 successively with \(x, x(1+1/\log ^3 x), x(1+1/\log ^3 x)^2\), keeping track of the resulting primes and bounds. For \(n\) large and \(x = \sqrt{n}\), we have \(\sqrt{n} {\lt} p_1\) as soon as the first interval lies strictly above \(\sqrt{n}\); this can be enforced by taking \(n\) large enough.

Apply Theorem 50 with suitable values of \(x\) slightly below \(n\), e.g. \(x = n(1+1/\log ^3\sqrt{n})^{-i}\), again keeping track of the intervals. For \(n\) large enough, these intervals lie in \((\sqrt{n},n)\) and contain primes \(q_i\) with the desired ordering.

By Lemma 140, each \(q_i\) is at least

for suitable indices \(j\), so \(1/q_i\) is bounded above by

after reindexing. Multiplying the three inequalities gives 6.

By Lemma 139, \(p_i \le \sqrt{n} (1+1/\log ^3\sqrt{n})^i\). Hence

From these one gets

and hence

Taking \(1+\cdot \) and multiplying over \(i=1,2,3\) gives 7.

We have \(p_i \le \sqrt{n} (1+1/\log ^3 \sqrt{n})^i\), so

Similarly, \(q_i \ge n(1+1/\log ^3 \sqrt{n})^{-i}\), so

Combining,

This implies 8.

This is a straightforward calculus and monotonicity check: the left-hand sides are decreasing in \(n\) for \(n \ge X_0^2\), and equality (or the claimed upper bound) holds at \(n=X_0^2\). One can verify numerically or symbolically.

Expand each finite product as a polynomial in \(\varepsilon \), estimate the coefficients using the bounds in Lemma 144, and bound the tails by simple inequalities such as

for small \(\varepsilon \). All coefficients can be bounded numerically in a rigorous way; this step is a finite computation that can be checked mechanically.

This is equivalent to

or

Factor out \(\varepsilon \) and use that \(0{\lt}\varepsilon \le 1/89693^2\) to check that the resulting quadratic in \(\varepsilon \) is nonnegative on this interval. Again, this is a finite computation that can be verified mechanically.

Combine Lemma 141, Lemma 142, and Lemma 143. Then apply Lemma 144 and set \(\varepsilon = 1/n\). The products are bounded by the expressions in Lemma 145, and the inequality in Lemma 146 then ensures that 5 holds.

By Proposition 8, there exist primes \(p_1,p_2,p_3,q_1,q_2,q_3\) satisfying the hypotheses of Theorem 51. Hence \(L_n\) is not highly abundant.