1 Introduction and Main Results

Let p(n) be the number of partitions of n [2, 3, 17]. It is well known that this arithmetic function increases strictly:

$$\begin{aligned} p(1) \, p(n) < p(n+1) \end{aligned}$$
(1.1)

for all \(n \in \mathbb {N}\), since every partition of n can be lifted to a partition of \(n+1\). Bessenrodt and Ono [5] discovered that (1.1) is actually an exception in the context of more general products \(p(a)\,p(b)\). Recently, DeSalvo and Pak [8] proved that the sequence \(\{p(n)\}\) is log-concave for \(n>25\).

Theorem 1.1

(Bessenrodt–Ono 2016) Let a, b be natural numbers. Let \(a+b >9\), then

$$\begin{aligned} p(a) \, p(b) > p(a+b). \end{aligned}$$
(1.2)

Bessenrodt and Ono provided proof, based on a theorem of Rademacher [18] and Lehmer [15]. They speculated at the end of their paper that a combinatorial proof could be possible. Shortly after their paper was published, Alanazi, Gagola, and Munagi [1] found such a proof. Chern, Fu, and Tang [6] generalized Bessenrodt and Ono’s theorem to k-colored partitions \(p_{-k}(n)\) of n.

Theorem 1.2

(Chern, Fu, Tang 2018) Let a, b, k be natural numbers. Let \(k>1\), then

$$\begin{aligned} p_{-k}(a) \, p_{-k}(b) > p_{-k}(a+b), \end{aligned}$$
(1.3)

except for \((a,b,k) \in \left\{ (1,1,2), (1,2,2), (2,1,2), (1,3,2), (3,1,2), (1,1,3)\right\} \).

Let \(P_n(x)\) be the unique polynomial of degree n satisfying \(P_n(k)=p_{-k}(n)\) for all \(k \in \mathbb {N}\) (see [9]). In this paper we prove the following results:

Theorem 1.3

Let \(n \in \mathbb {N}\) and \(x \in \mathbb {R}\) with \(x \ge 1\). Then

$$\begin{aligned} P_n(x)< P_{n+1}(x) \quad \text {and}\quad 1 \le P_n'(x)< P_{n+1}'(x). \end{aligned}$$
(1.4)

Moreover, let \(n+1\) be an odd prime number. Then, there exists \(x_n \in (0,1)\), such that

$$\begin{aligned} P_{n+1}(x_n)<P_n(x_n). \end{aligned}$$
(1.5)

Our main result is the following extension of the Bessenrodt–Ono type inequality.

Theorem 1.4

Let \(a,b \in \mathbb {N}\), \(a+b>2\), and \(x >2\). Then

$$\begin{aligned} P_a(x) \, P_b(x) > P_{a+b}(x). \end{aligned}$$
(1.6)

The case \(x=2\) is true for \(a+b>4\).

This gives a precise answer to a conjecture stated in [11]. The theorem is proven by induction. The proof uses a special formula for the derivative of \(P_n(x)\), the inequality (1.3) for \(k=2\), a bound of Lehmer (5.2), and Theorem 1.3. The inequality (1.3) for \(k=2\) is proven by Chern, Fu, and Tang [6].

The values in Table 1 for \(1\le a,b\le 10\) imply that inequality (1.6) can also hold for values of x smaller than 2. If we assume \(a\ge b\) then except for \(b=1\) and \(a=1,2,3\) all values are less than 2. For values of x larger than the numbers (rounded to two decimal places) in position \(\left( a,b\right) \) inequality (1.6) holds.

Bessenrodt–Ono-type inequalities also appeared in works by Beckwith and Bessenrodt [4] on k-regular partitions and by Hou and Jagadeesan [12] on the numbers of partitions with ranks in a given residue class modulo \(t=3\). Recently, Males [16] obtained results for general t. Dawsey and Masri [7] obtained new results for Andrews spt-function. It is very likely that some of these recent results can be extended to an inequality of certain polynomials.

This paper is organized as follows. In Sect. 2, we introduce the polynomials \(P_n(x)\) and provide basic properties. In Sect. 3, we give numerical evidence for Theorem 1.3 and Theorem 1.4. In Sect. 4, we prove Theorem 1.3 and in Sect. 5, we prove Theorem 1.4.

2 Partitions and Polynomials

A partition \(\lambda \) of a positive integer n is any non-increasing sequence \(\lambda _1,\lambda _2, \ldots , \lambda _d\) of positive integers, whose sum is n. The \(\lambda _i\) denote the parts of the partition. The number of partitions of n is denoted by p(n) (see [2, 17]).

Example

The partitions of \(n=4\) and \(n=5\) are

$$\begin{aligned} 4= & {} 3+1=2+2=2+1+1=1+1+1+1\\ 5= & {} 4+1=3+2=3+1+1=2+2+1=2+1+1+1=1+1+1+1+1. \end{aligned}$$

Hence, \(p(4)=5\) and \(p(5)=7\). Note that p(200) is already equal to 3972999029388.

A partition is called a k-colored partition of n if each part can appear in k colors. Let \(p_{-k}(n)\) denote the number of k-colored partitions of n (see [6]). Note that \(p_{-1}(n)= p(n)\) and \(p_{-k}(n) < p_{-(k+1)}(n)\). The generating function of \(p_{-k}(n)\) is given by

$$\begin{aligned} \sum _{n=0}^{\infty } p_{-k}(n) \, q^n = \frac{1}{ \prod _{n=1}^{\infty } \left( 1 - q^n \right) ^k } = \frac{1}{(q;q)_{\infty }^k} \qquad (k \in \mathbb {N}). \end{aligned}$$
(2.1)

Here, \((a;q)_{\infty }= \prod _{n=0}^{\infty } (1 - a \, q^n)\) is the q-Pochhammer symbol.

Definition

We define recursively a family of polynomials \(P_n(x)\). Let \(\sigma (n) := \sum _{d \mid n} d\) denote the sum of divisors of n. Then let \(P_0(x):=1\) and

$$\begin{aligned} P_n(x) := \frac{x}{n} \sum _{k=1}^{n} \sigma (k) \, P_{n-k}(x). \end{aligned}$$
(2.2)

It is known that \(P_n(k) = p_{-k}(n)\) for \(k \in \mathbb {N}\) (see [9] and compare the generating functions of \(P_{n}\left( k\right) \) and \(p_{-k}\left( n\right) \)). Let \(p_{-k}(0):=1\). If we put \(q:= \mathrm {e}^{2 \pi \mathrm {i} \tau }\) with \(\tau \) in the upper complex half-plane, then

$$\begin{aligned} \sum _{n=0}^{\infty } P_n(z) \,\, q^n = \prod _{n=1}^{\infty } \left( 1 - q^n \right) ^{-z} \qquad (z \in \mathbb {C}). \end{aligned}$$

We have \(P_0(x)=1\), \(P_1(x) = x\), \(P_2(x) = x/2 \, (x+3)\), \(P_3(x) = x/6 \, (x^2 + 9x +8)\). In addition, we have \(P_n(x) = x/n! \, \cdot \, {\widetilde{P}}_n(x)\), where \({\widetilde{P}}_n(x)\) is a monic polynomial of degree \(n-1\) with positive integer coefficients. There is a useful formula for the derivatives \(P_n'(x)\) for all \(n \in \mathbb {N}\) [10]:

$$\begin{aligned} P_n'(x) = \sum _{k=1}^n \frac{\sigma (k)}{k} \, P_{n-k}(x). \end{aligned}$$
(2.3)

3 Numerical Evidence for Theorems 1.3 and 1.4

3.1 The Difference Functions

In this section, we study the difference function

$$\begin{aligned} \Delta _n(x):= P_{n+1}(x) - P_n(x) \end{aligned}$$
(3.1)

for \(n\in \mathbb {N}\). The results will be utilized to prove Theorem 1.3. We first observe that

$$\begin{aligned} \lim _{x \rightarrow \infty } \Delta _n(x) = + \infty . \end{aligned}$$
(3.2)

This is true since \(P_n(x)\) is a polynomial of degree n of leading coefficient 1/n!. We also deduce that \(\Delta _n(0)=0\) for \(n\ge 1\). We are especially interested in the largest non-negative real root \(x_n\) of \(\Delta _n(x)\) (compare Fig. 1), since \(P_{n+1}(x) > P_n(x)\) for \(x > x_n\).

By (3.2) and continuity for all values of x larger than this root the difference function is positive. The real roots of \(\Delta _1(x)\) are \(\{-1,0\}\) and the real roots of \( \Delta _2(x)\) are

$$\begin{aligned} \{-3-\sqrt{10}, \,0, \, -3+ \sqrt{10} \}. \end{aligned}$$

We can see already that \(\Delta _n(x)\) is not always positive, here \( \Delta _2 ( x) <0\) for

$$\begin{aligned} 0< x < -3+ \sqrt{10}. \end{aligned}$$
Fig. 1
figure 1

Roots of \(\Delta _{n}(x)\) with positive real part. The real part is displayed

Full size image

3.2 The Multiplicative Differences Functions

Let \(a,b \in \mathbb {N}\). The Bessenrodt–Ono inequality \(p(a) \, p(b) > p(a+b)\) is always satisfied for all \(a,b \ge 2\) and \(a+b >9\). Since the inequality is symmetric in a and b we assume \(a \ge b\). It was also shown [5] that there is equality for \((a,b) \in \left\{ (6,2), (7,2), (4,3)\right\} \). The inequality fails completely for \(b=1\) and

$$\begin{aligned} (a,b) \in \left\{ (2,2),(3,2), (4,2), (5,2), (3,3),(5,3)\right\} , \end{aligned}$$

while it is true for the remaining cases \((a,b)\in \left\{ (4,4),(5,4),\left( 6,3\right) \right\} \).

The BO for 2-colored partitions is true for all \(a,b \in \mathbb {N}\) except for \((a,b)=(1,1)\), where \(p_{-2} (1) \, p_{-2}(1) < p_{-2}(2)\). Let \(a \ge b\). Then, we have equality for \((a,b) \in \{ (2,1),(3,1)\}\). The BO for 3-colored partitions holds for all \(a,b \in \mathbb {N}\) except for \((a,b)=(1,1)\), where we have equality. If \(k \ge 4\), then BO is fulfilled for all \(a,b \in \mathbb {N}\) (see [6]).

Let \(P_{a,b}(x):= P_a(x) \, P_b(x) - P_{a+b}(x)\). Then \(P_{a,b}(0)= 0\), \(P_{a,b}'(0) = - \sigma (a+b)/(a+b)\) and

$$\begin{aligned} \lim _{ x \rightarrow \infty } P_{a,b}(x) = \infty . \end{aligned}$$

In contrast to \(\Delta _n(x)\), the polynomials \(P_{a,b}(x)\) appear to have only one root \(x_{a,b}\) with a positive real part. Table 1 records these roots for \(1 \le a,b \le 10\). The root distribution in Table 1 explains all exceptions in the papers [5] and [6]. In Fig. 2 we have displayed the single positive root \(x_{a,1}\) of \(P_{a,1}(x)\) for \(1 \le a \le 100\).

Table 1 Positive real roots of \(P_{a,b}(x)\)
Full size table

Conjecture

It seems that, in general, \(P_{a,1}(x)\) has exactly one positive real root (and no non-real roots) and that the limit exists and is equal to 1.

4 Proof of Theorem 1.3

Proof of Theorem 1.3

We prove the first part of Theorem 1.3 by induction.

Claim: \(P_{n+1}(x) > P_n(x)\) for all \(n \in \mathbb {N}\) and \(x \ge 1\). Let \(n=1\). We have \(\Delta _1(x)> 0\) for all \(x>0\). Suppose that \(\Delta _m(x)>0\) is true for all real numbers \(x \ge 1\) and integers \(1 \le m \le n-1\). Using (2.3) and the induction hypothesis, we obtain for \(P_{n+1}'(x)\) the strict lower bound:

$$\begin{aligned} \sum _{k=1}^n \frac{\sigma (k)}{k} \, P_{n+1-k}(x) \ge \sum _{k=1}^n \frac{\sigma (k)}{k} \, P_{n-k}(x). \end{aligned}$$
(4.1)

Hence, \(P_{n+1}'(x) > P_{n}'(x)\). Property (1.1) for the partition function provides

$$\begin{aligned} P_{n+1}(1) = p(n+1)> p(n) = P_n (1). \end{aligned}$$
(4.2)

As a corollary, we obtain that \(\Delta _n'(x) >0\).

We finally prove (1.5). We have already observed \(\Delta _{2}\left( x\right) <0\) for \(0<x<\sqrt{10}-3\) in Sect. 3.1. Since \(\Delta _n(0)=0\) and

$$\begin{aligned} \lim _{x \rightarrow \infty } \Delta _n(x) = + \infty \end{aligned}$$

it is sufficient to show that \(\Delta _n'(0) <0\) for \(n+1\) an odd prime. We have

$$\begin{aligned} \Delta _n'(0) = \frac{\sigma (n+1)}{n+1} - \frac{\sigma (n)}{n} = \frac{n+2}{n+1} - \frac{\sigma (n)}{n} <0 \end{aligned}$$
(4.3)

using that \(\sigma (n) > n+1\).

Fig. 2
figure 2

Positive real roots of \(P_{a,1}(x)\)

Full size image

5 Proof of Theorem 1.4

5.1 Lehmer’s Bound

Let \(\mu (n):= \frac{\pi }{6} \sqrt{24n-1}\). Rademacher [18] proved the formula:

$$\begin{aligned} p(n) = \frac{\sqrt{12}}{24n-1} \sum _{k=1}^{N} A_k^{*}(n) \, \left( \left( 1-\frac{k}{\mu }\right) e^{\mu /k } + \left( 1+ \frac{k}{\mu }\right) e^{- \mu /k } \right) + R_2(n,N). \end{aligned}$$

Here, \(A_k^{*}(n) = \frac{1}{\sqrt{k}} A_k(n)\) are real numbers, where \(A_k(n)\) is a complicated sum of 24kth roots of unity. Lehmer [13,14,15] obtained the estimate

$$\begin{aligned} \vert R_2(n,N) \vert < \frac{\pi ^2 N^{-2/3}}{\sqrt{3}} \left( \left( \frac{N}{\mu }\right) ^{3} \, \text {sinh}\left( \frac{\mu }{N}\right) + \frac{1}{6} - \left( \frac{N}{\mu }\right) ^{2} \right) . \end{aligned}$$
(5.1)

for all \(n, N \in \mathbb {N}\). DeSalvo and Pak [8] recently utilized the case \(N=2\) and proved that p(n) is log-concave for all \(n>25\). They proved two of Chen’s conjectures. For our purpose the case \(N=1\), which was studied by Bessenrodt and Ono [5], is more convenient. They obtained:

$$\begin{aligned} \frac{\sqrt{3}}{12m} \left( 1 - \frac{1}{\sqrt{m}}\right) \mathrm {e}^{\frac{\pi }{6}\sqrt{ 24m-1}}< P_{m}\left( 1\right) < \frac{\sqrt{3}}{12m} \left( 1 + \frac{1}{\sqrt{m}}\right) \mathrm {e}^{\frac{\pi }{6}\sqrt{24m-1}}. \end{aligned}$$
(5.2)

This is the effective estimate for \(P_{m}\left( 1\right) \) that we use throughout our proof.

5.2 Proof of Theorem 1.4 When \(b = 1\)

In the following, we utilize the well-known upper bound:

$$\begin{aligned} \sigma \left( m\right) \le m \left( 1+\ln \left( m\right) \right) . \end{aligned}$$
(5.3)

Proposition 5.1

Let \(n \in \mathbb {N}\) and \(x \in \mathbb {R}\). The inequality

$$\begin{aligned} x \, P_{n}\left( x\right) >P_{n+1}\left( x\right) \end{aligned}$$
(5.4)

holds for all \(x>2\) and \(n\ge 2\). In the case \(n=1\) it holds for \(x>3\).

Proof

The case \(n=1\) is easy to see, since \(P_{1,1}(x) = x/2 \, (x-3)\). We prove the proposition by induction on n. Let \(n=2,3 \). Then \(P_{2,1}(x) = x/3 \, (x^2-4)>0\) for \(x >2\) and \(P_{3,1}\left( x\right) =\frac{x}{8}\,\left( x-2\right) \left( x+1\right) \left( x+7\right) >0\) for \(x>2\). Suppose that \(n\ge 4 \) and that \(P_{m,1}(x) >0\) for \(2 \le m \le n-1\) and \(x>2\). It is sufficient to prove that \(\frac{\mathrm {d}}{\mathrm {d}x} P_{n,1}(x) > 0\), since we already know that \(P_{n,1}(2) \ge 0\) ([6], see Theorem 1.2). The derivative of \(P_{n,1}(x)\) is equal to

$$\begin{aligned}&P_{n}\left( x\right) +\sum _{k=1}^{n}\frac{\sigma \left( k\right) }{k}xP_{n-k}\left( x\right) -\sum _{k=1}^{n+1}\frac{\sigma \left( k\right) }{k}P_{n+1-k}\left( x\right) \\&\quad >P_{n}\left( x\right) +\frac{\sigma \left( n-1\right) }{n-1}\left( \left( x-3\right) x/2\right) -\frac{\sigma \left( n+1\right) }{n+1} \\&\quad \ge P_{n}\left( x\right) -\left( 1+\ln \left( n-1\right) \right) \frac{9}{8}-\left( 1+\ln \left( n+1\right) \right) \\&\quad \ge P_{n}\left( x\right) -\frac{17}{8}\left( 1+\ln \left( 2n\right) \right) . \end{aligned}$$

We recall that \(P_n(x)\) has non-negative coefficients. It is, therefore, sufficient to show that

$$\begin{aligned} P_{n}\left( 2\right) \ge \frac{17}{8}\left( 1+\ln \left( 2n\right) \right) \end{aligned}$$
(5.5)

for \(n\ge 4 \). Since \(P_n(2)> P_n(1)\), we can now use (5.2). Suppose \(n\ge 49\). Then

$$\begin{aligned} P_{n}\left( 1\right)> & {} \frac{\sqrt{3}}{12n}\left( 1-\frac{1}{\sqrt{n}}\right) \mathrm {e}^{\frac{\pi }{6}\sqrt{24n-1}} \\> & {} \frac{\sqrt{3}}{4!12n} \left( 1 - \frac{1}{\sqrt{n}}\right) \left( \frac{\pi }{6} \sqrt{24n-1}\right) ^{4} \\= & {} \frac{\sqrt{3}}{4!12}\frac{24n-1}{n}\left( 1-\frac{1}{\sqrt{n}}\right) \left( \frac{\pi }{6}\right) ^{4}\left( 24n-1\right) \\> & {} \frac{23\sqrt{3}}{4!12}\frac{6}{7}\left( \frac{\pi }{6}\right) ^{4}\left( 24n-1\right) =:f\left( n\right) . \end{aligned}$$

For

$$\begin{aligned} n\ge 58>\left( \frac{23\sqrt{3}}{4!12}\frac{6}{7}\left( \frac{ \pi }{6}\right) ^{4}+\frac{17}{8}\ln \left( 98\right) \right) \Bigg {/}\left( \frac{23\sqrt{3}}{12}\frac{6}{7}\left( \frac{\pi }{6}\right) ^{4}-\frac{17}{392}\right) \end{aligned}$$

we see that \( 17/8 \left( 1+\ln \left( 2n\right) \right) \le 17/8 \left( \ln \left( 98\right) +\frac{n}{49} \right) \le f\left( n\right) \). Thus for \(n\ge 58\) it holds that \(P_{n}\left( x\right)>P_{n}\left( 1\right) > 17/8 \left( \ln \left( 98\right) +\frac{n}{49} \right) \ge 17/8 \left( 1+\ln \left( 2n\right) \right) \). It remains to check \(P_{n}\left( 2\right) >\frac{17}{8}\left( 1+\ln \left( 2n\right) \right) \) for \(4 \le n\le 57\). In this case we have \(\left( 1+\ln \left( 2n\right) \right) <6\). Since \(P_{n}\left( 2\right) \) increases monotonously in n by Theorem 1.3 we also have \(\frac{17}{8}\left( 1+\ln \left( 2n\right) \right) < 20=P_{4 }\left( 2\right) \le P_{n}\left( 2\right) \) for \(4 \le n\le 57\). \(\square \)

We deduce from the proof of Proposition 5.1 the following property.

Corollary 5.2

Let \(n \ge 2\) and \(x>2\) or \( n \ge 4\) and \(x \ge 1\). Then

$$\begin{aligned} P_n(x) - \left( 1+\ln \left( 2n\right) \right) >0. \end{aligned}$$
(5.6)

5.3 Proof of Theorem 1.4 When \(b > 1\)

Proof of Theorem 1.4

We show (1.6) by induction on \(n=a+b\). For \(n=a+b=3\) we have \(P_{1}\left( x\right) P_{2}\left( x\right) -P_{3}\left( x\right) =\frac{x}{3}\left( x^{2}-4\right) >0\) for all \(x>2\).

Suppose \(n\ge 4\) and \(P_{A}\left( x\right) P_{B}\left( x\right) >P_{A+B}\left( x\right) \) for all \(3\le A+B\le n-1=a+b-1\) and \(x>2\). Without loss of generality we assume \(a\ge b\ge 2\). (The case \(b=1\) was proved in Proposition  5.1.)

We have \(P_{a}\left( 2\right) P_{b}\left( 2\right) \ge P_{a+b}\left( 2\right) \) for \(a+b\ge 3\) by Theorem 1.2 of [6]. If we can now show that

$$\begin{aligned} \frac{\mathrm {d}}{\mathrm {d}x}\left( P_{a}\left( x\right) P_{b}\left( x\right) \right) >P_{a+b}^{\prime }\left( x\right) \end{aligned}$$
(5.7)

the proof is completed, as this implies \(P_{a}\left( x\right) P_{b}\left( x\right) >P_{a+b}\left( x\right) \) for all \(x>2\).

Note that \(P_{A}\left( x\right) P_{0}\left( x\right) =P_{A}\left( x\right) \). Thus,

$$\begin{aligned}&P_{a}^{\prime }\left( x\right) P_{b}\left( x\right) +P_{a}\left( x\right) P_{b}^{\prime }\left( x\right) -P_{a+b}^{\prime }\left( x\right) \\&\quad =\sum _{k=1}^{a}\frac{\sigma \left( k\right) }{k}P_{a-k}\left( x\right) P_{b}\left( x\right) +P_{a}\left( x\right) \sum _{k=1}^{b}\frac{\sigma \left( k\right) }{k}P_{b-k}\left( x\right) -\sum _{k=1}^{a+b}\frac{\sigma \left( k\right) }{k}P_{a+b-k}\left( x\right) \\&\quad >\sum _{k=1}^{a}\frac{\sigma \left( k\right) }{k}P_{a+b-k}\left( x\right) +\sum _{k=1}^{b}\frac{\sigma \left( k\right) }{k}P_{a+b-k}\left( x\right) -\sum _{k=1}^{a+b}\frac{\sigma \left( k\right) }{k}P_{a+b-k}\left( x\right) \\&\quad =\sum _{k=1}^{b}\frac{\sigma \left( k\right) }{k}P_{a+b-k}\left( x\right) -\frac{\sigma \left( k+a\right) }{k+a}P_{b-k}\left( x\right) \\&\quad \ge \sum _{k=1}^{b}P_{a+b-k}\left( x\right) -\left( 1+\ln \left( 2a\right) \right) P_{b-k}\left( x\right) . \end{aligned}$$

We now consider

$$\begin{aligned} P_{a+b-k}\left( x\right) -\left( 1+\ln \left( 2a\right) \right) P_{b-k}\left( x\right) \end{aligned}$$
(5.8)

for each k separately. From Theorem 1.3, we know that \(P_{a+b-k}\left( x\right) \) increases faster than \(P_{b-k}\left( x\right) \) for \(x\ge 1\). Hence, to show that (5.8) is positive, it is enough to show this for \(1\le x\le 2\).

Using (5.2) for \(k<b\)

$$\begin{aligned}&P_{a+b-k}\left( 1\right) -\left( 1+\ln \left( 2a\right) \right) P_{b-k}\left( 1\right) \\&\quad >\frac{\sqrt{3}}{12\left( a+b-k\right) }\left( 1-\frac{1}{\sqrt{a+b-k}}\right) \mathrm {e}^{\frac{\pi }{6}\sqrt{24\left( a+b-k\right) -1}} \\&\qquad -\left( 1+\ln \left( 2a\right) \right) \frac{\sqrt{3}}{12\left( b-k\right) }\left( 1+\frac{1}{\sqrt{b-k}}\right) \mathrm {e}^{\frac{\pi }{6}\sqrt{24\left( b-k\right) -1}} \end{aligned}$$

and the last is positive if and only if

$$\begin{aligned} \mathrm {e}^{\frac{\pi }{6}\left( \sqrt{24\left( a+b-k\right) -1}-\sqrt{24\left( b-k\right) -1}\right) } >\left( 1+\ln \left( 2a\right) \right) \frac{a+b-k}{b-k}\frac{1+\frac{1}{\sqrt{b-k}}}{1-\frac{1}{\sqrt{a+b-k}}}. \end{aligned}$$
(5.9)

Now

$$\begin{aligned}&\sqrt{24\left( a+b-k\right) -1}-\sqrt{24\left( b-k\right) -1} \\&\quad =\frac{24a}{\sqrt{24\left( a+b-k\right) -1}+\sqrt{24\left( b-k\right) -1}} \\&\quad \ge \frac{24a}{\sqrt{24\left( 2a-1\right) -1}+\sqrt{24\left( a-1\right) -1}} \\&\quad >\frac{24a}{7\sqrt{a}+5\sqrt{a}}=2\sqrt{a}. \end{aligned}$$

On the other hand

$$\begin{aligned} \left( 1+\ln \left( 2a\right) \right) \frac{a+b-k}{b-k}\frac{1+\frac{1}{\sqrt{b-k}}}{1-\frac{1}{\sqrt{a+b-k}}} <\left( 1+\ln \left( 2a\right) \right) \left( 1+a\right) \frac{2}{1-\frac{1}{\sqrt{a}}}. \end{aligned}$$

If a fulfills

$$\begin{aligned} \mathrm {e}^{\pi \sqrt{a}/3}>\left( 1+\ln \left( 2a\right) \right) \left( 1+a\right) \frac{2}{1-\frac{1}{\sqrt{a}}} \end{aligned}$$
(5.10)

then a also fulfills (5.9). We show (5.10) for \(a\ge 94\) in Lemma 5.3 below. This implies that (5.8) is positive for \(x=1\). Theorem 1.3 then implies that (5.8) is positive for \(x\ge 1\).

We used PARI/GP to check that Eq. (5.8) with \(x=2\) is positive for \(1\le k<b\le a\le 93\). Again Theorem 1.3 implies (5.8) is positive for \(x\ge 2\).

What remains to be considered is (5.8) for \(k=b\). This follows from Corollary 5.2 setting \(n=a\).

Thus, for all \(a\ge b\ge k\ge 1\) the value of (5.8) is positive. Hence, also

$$\begin{aligned} P_{a}^{\prime }\left( x\right) P_{b}\left( x\right) +P_{a}\left( x\right) P_{b}^{\prime }\left( x\right) >P_{a+b}^{\prime } \left( x\right) \end{aligned}$$

for \(x>2\). This completes the induction step, as explained in the beginning of the proof. \(\square \)

In the following, we provide a proof of (5.10) for \(a\ge 94\). With more work and tighter bounds, it is possible to show (5.10) for \(a\ge 34\).

Lemma 5.3

For \(a\ge 94\), Eq. (5.10) holds, that is

$$\begin{aligned} \mathrm {e}^{\pi \sqrt{a}/3}>\left( 1+\ln \left( 2a\right) \right) \left( 1+a\right) \frac{2}{1-\frac{1}{\sqrt{a}}}. \end{aligned}$$

Proof

Since \(a>0\)

$$\begin{aligned} \mathrm {e}^{\pi \sqrt{a}/3}>\frac{1}{6!}\left( \frac{\pi }{3}\sqrt{a}\right) ^{6}=\frac{1}{6!}\left( \frac{\pi }{3}\right) ^{6}a^{3}. \end{aligned}$$

Since \(a\ge 81\), we have

$$\begin{aligned} \left( 1+\ln \left( 2a\right) \right) \left( 1+a\right) \frac{2}{1-\frac{1}{\sqrt{a}}}\le \left( \frac{a}{81}+\ln \left( 162\right) \right) \left( 1+a\right) \frac{18}{8}. \end{aligned}$$

Thus, (5.10) is fulfilled if

$$\begin{aligned} \frac{1}{6!}\left( \frac{\pi }{3}\right) ^{6}a\ge \left( \frac{1}{81}+\frac{\ln \left( 162\right) }{81}\right) \left( \frac{1}{81}+1\right) \frac{18}{8}\ge \left( \frac{1}{81}+\frac{\ln \left( 162\right) }{a}\right) \left( \frac{1}{a}+1\right) \frac{9}{4} \end{aligned}$$

and this is fulfilled for \(a\ge 94>\left( \left( \frac{1}{81}+\frac{\ln \left( 162\right) }{81}\right) \left( \frac{1}{81}+1\right) \frac{9}{4}\right) /\left( \frac{1}{6!}\left( \frac{\pi }{3}\right) ^{6}\right) \). \(\square \)