From (1.4), we see that \(d_{m}(m)=2^{-m}{2m \choose m} \), which implies that (1.7) holds.

By (1.4),

$$d_{m}(m+1)= \frac{(2m+3)(2m+1)}{2(m+1)} 2^{-m}{2m\choose m}, $$

which yields

$$d_{i}^{2}(i+1)>d_{i-1}(i)d_{i+1}(i+2). $$

Therefore, (1.6) holds when \(m=1\).

Hence, in the following, we always assume that \(m\geq2\) and \(i\geq1\). We first recall the following three recurrence relations derived by Kauers and Paule [3]:

$$\begin{aligned}& d_{i}(m+1)=\frac{m+i}{m+1}d_{i-1}(m)+\frac{(4m+2i+3)}{2(m+1)}d_{i}(m), \quad 0 \leq i \leq m+1, \end{aligned}$$

(2.1)

$$\begin{aligned}& d_{i}(m+1)=\frac{(4m-2i+3)(m+i+1)}{2(m+1)(m+1-i)}d_{i}(m) \\& \hphantom{d_{i}(m+1)={}}{}-\frac{i(i+1)}{(m+1)(m+1-i)}d_{i+1}(m), \quad 0 \leq i \leq m, \end{aligned}$$

(2.2)

and

$$\begin{aligned} d_{i}(m+2) =&\frac{-4i^{2}+8m^{2}+24m+19}{2(m+2-i) (m+2)}d_{i}(m+1) \\ &{} -\frac{(m+i+1)(4m+3)(4m+5)}{ 4(m+2-i)(m+1)(m+2)}d_{i}(m),\quad 0 \leq i \leq m+1. \end{aligned}$$

(2.3)

Now we represent the difference \(d_{i}^{2}(m+i)-d_{i-1}(m+i-1)d_{i+1}(m+i+1)\) in terms of \(d_{i}(m+i)\) and \(d_{i}(m+i+1)\). Thanks to (2.1), (2.2) and (2.3),

$$\begin{aligned}& d_{i}^{2}(m+i)-d_{i-1}(m+i-1) d_{i+1}(m+i+1) \\& \quad = A d_{i}^{2}(m+i+1) +B d_{i}(m+i+1)d_{i}(m+i)+C d_{i}^{2}(m+i), \end{aligned}$$

(2.4)

where

$$\begin{aligned}& A =\frac{(4m+6i+5) (m+1+i)(m+i)(m+1)^{2}(4m+6i-1)}{ (i+1)i(4m+4i+1)(4m+4i-1)(m+2i)(m+2i-1)}, \end{aligned}$$

(2.5)

$$\begin{aligned}& B =-\frac{(m+1)(m+i)D}{ (i+1)i(4m+4i+1)(4m+4i-1)(m+2i)(m+2i-1)}, \end{aligned}$$

(2.6)

$$\begin{aligned}& C =\frac{E}{4(m+2i-1)(m+2i)(4m+4i-1) (4m+4i+1)(m+1+i)i(i+1)} \end{aligned}$$

(2.7)

with

$$\begin{aligned}& D= -15+400mi+35i+13m+140m^{2}+292i^{2} +864mi^{2}+688m^{2}i \\& \hphantom{D={}}{}+176m^{3}+336i^{3}+64m^{4}+72i^{4} +320m^{3}i +560m^{2}i^{2}+384mi^{3}, \end{aligned}$$

(2.8)

$$\begin{aligned}& E= -68mi-45i-45m-66m^{2}+2i^{2} +2\text{,}614mi^{2}+1 \text{,}901m^{2}i+451m^{3} +1\text{,}164i^{3} \\& \hphantom{E={}}{}+1\text{,}560m^{4}+3\text{,}320i^{4} +7 \text{,}732m^{3}i+14\text{,}176m^{2}i^{2}+11 \text{,}328mi^{3}+1\text{,}152i^{6} +1\text{,}984m^{5} \\& \hphantom{E={}}{}+3\text{,}392i^{5}+11\text{,}888m^{4}i +16 \text{,}856i^{4}m +27\text{,}772m^{3}i^{2}+31 \text{,}332m^{2}i^{3} +8\text{,}128m^{5}i \\& \hphantom{E={}}{}+23\text{,}040m^{4}i^{2}+9 \text{,}216i^{5}m +33\text{,}216m^{3}i^{3}+25 \text{,}216m^{2}i^{4} +6\text{,}720m^{5}i^{2}+11 \text{,}584m^{4}i^{3} \\& \hphantom{E={}}{}+1\text{,}152i^{6}m+11\text{,}072m^{3}i^{4}+5 \text{,}568m^{2}i^{5} +2\text{,}048m^{6}i +1 \text{,}152m^{6}+256m^{7}. \end{aligned}$$

(2.9)

It is easy to check that

$$ \Delta =B^{2}-4AC=\frac{(m+1)^{2} (m+i)F}{i(i+1)^{2}(4i+4m+1)^{2} (4i+4m-1)^{2}(2i+m)^{2}(2i+m-1)^{2}}, $$

where

$$\begin{aligned} F =& 5\text{,}184i^{8}+19\text{,}008i^{7}m+27 \text{,}648i^{6}m^{2} +19\text{,}968i^{5}m^{3}+7 \text{,}168i^{4}m^{4} +1\text{,}024i^{3}m^{5} \\ &{}+6\text{,}912i^{7}+16\text{,}128i^{6}m +768i^{5}m^{2}-33 \text{,}024i^{4}m^{3}-44\text{,}288i^{3}m^{4} -26\text{,}880i^{2}m^{5} \\ &{}-8\text{,}192im^{6}-1\text{,}024m^{7}+5\text{,}184i^{6}+13 \text{,}920i^{5}m+9\text{,}584i^{4}m^{2} -5 \text{,}936i^{3}m^{3} \\ &{}-11\text{,}648i^{2}m^{4}-5 \text{,}888im^{5}-1\text{,}024m^{6}+6\text{,}096i^{5} +23 \text{,}488i^{4}m+35\text{,}600i^{3}m^{2} \\ &{}+26 \text{,}512i^{2}m^{3} +9\text{,}728im^{4}+1 \text{,}408m^{5}+2\text{,}000i^{4}+7\text{,}232i^{3}m+9 \text{,}536i^{2}m^{2} \\ &{}+5\text{,}360im^{3}+1 \text{,}088m^{4}-1\text{,}048i^{3}-2\text{,}336i^{2}m-1\text{,}728im^{2}-404m^{3} \\ &{}-143i^{2} -175im-64m^{2}+40i+20m. \end{aligned}$$

Note that *A* is positive. Hence, in order to prove that the right-hand side of (2.4) is positive, it suffices to prove that when Δ is nonnegative,

$$ \frac{d_{i}(m+i+1)}{d_{i}(m+i)}>\frac{-B+\sqrt{\Delta}}{2A}. $$

(2.10)

Therefore, in the following, we assume that \(\Delta\geq0\).

Recall that Kauers and Paule [3] proved the following inequality:

$$\frac{d_{i}(m+1)}{d_{i}(m)}\geq\frac{4m^{2}+7m+i+3}{2(m+1)(m+1-i)},\quad 0\leq i \leq m. $$

Replacing *m* by \(m+i\), we see that

$$ \frac{d_{i}(m+i+1)}{d_{i}(m+i)}\geq \frac{4i^{2}+8im+4m^{2}+8i+7m+3}{2(m+1+i)(m+1)} , \quad i\geq0. $$

(2.11)

It is a routine to verify that

$$\begin{aligned}& \biggl(A \frac{4i^{2}+8im+4m^{2}+8i+7m+3}{ (m+1+i)(m+1)}+B \biggr)^{2} -\Delta \\& \quad = \frac{4(i+m)(m+1)^{2}(6i+4m+5) (6i+4m-1)G}{i(i+1)^{2}(4i+4m+1)^{2} (4i+4m-1)^{2} (2i+m)^{2}(2i+m-1)^{2}}, \end{aligned}$$

(2.12)

where

$$\begin{aligned} G =&28i^{4}m+108i^{3}m^{2} +144i^{2}m^{3}+80im^{4} +16m^{5}-32i^{4}-66i^{3}m \\ &{}-46i^{2}m^{2}-12im^{3}-32i^{3}-78i^{2}m -64im^{2}-17m^{3}+2i^{2}+2im+2i+m. \end{aligned}$$

Note that when \(m\geq2\) and \(i \geq1\), *G* is positive. Thus the right-hand side of (2.12) is positive. On the other hand,

$$\begin{aligned}& A \frac{4i^{2}+8im+4m^{2}+8i+7m+3}{(m+1+i)(m+1)}+B \\& \quad = \frac{(i+m)(m+1) (-3-12i+28im+48i^{2}+72i^{3}+32im^{2}+96i^{2}m)}{ (i+1)(4i+4m+1)(4i+4m-1)(2i+m)(2i+m-1)}, \end{aligned}$$

which is positive. Therefore, from (2.12), we have

$$A \frac{4i^{2}+8im+4m^{2}+8i+7m+3}{ (m+1+i)(m+1)}+B>\Delta, $$

which can be rewritten as

$$ \frac{4i^{2}+8im+4m^{2}+8i+7m+3}{2(m+1+i)(m+1)} >\frac{-B+\sqrt{\Delta}}{2A}. $$

(2.13)

From (2.11) and (2.13), we obtain (2.10) and this completes the proof.