# Three Cube Packing for All Dimensions
## 2 Results
The inequality \(\frac{\frac{1}{2k}+1}{0.74^{d}+\left(\left\{1-1\right\}^{d}+1\right)}\geq\frac {\sqrt{2}+1}{2^{1-\frac{1}{d}}+1}\) gives \(\frac{\sqrt{2}}{87}\Big{(}\left(\frac{37}{50}\right)^{d}+1\Big{)}\leq\frac{ \sqrt{2}}{87}\Big{(}\left(\frac{37}{50}\right)^{d}+1\Big{)}\leq\frac{\sqrt{2}} {87}\Big{(}\left(\frac{37}{50}\right)^{d}+1\Big{)}\leq\frac{\sqrt{2}}{87}\). If \(d\geq 11\), then the global maximum of \(G\) must occur on the interior of \(PC\). **Proof. ** Assume, for the sake of contradiction, that if \(d\geq 11\), then \(W_{PB}\geq G(P)\) at some \(k\in[0.74,1)\). Using the endpoints of the interval in \(W_{PB}\) we obtain an upper bound of \(W_{PB}\) on the interval. The inequality \(\frac{\frac{1}{2k}+1}{0.74^{d}+\left(\left\{1-1\right\}^{d}+1\right)}\geq\frac {\sqrt{2}+1}{2^{1-\frac{1}{d}}+1}\) gives \(\frac{37}{87}\Big{(}\left(\frac{37}{50}\right)^{d}+1\Big{)}\leq\frac{\sqrt{2}} {87}\Big{(}\left(\frac{37}{50}\right)^{d}+1\Big{)}\leq\frac{\sqrt{2}}{87}\). If \(d\geq 11\), then the global maximum of \(G\) must occur on the interior of \(PC\). **Proof. ** Assume, for the sake of contradiction, that if \(d\geq 11\), then \(W_{PB}\geq G(P)\) at some \(k\in[0.74,1)\). Using the endpoints of the interval in \(W_{PB}\) we obtain an upper bound of \(W_{PB}\) on the interval. The inequality \(\frac{\frac{1}{2k}+\(\left(\sqrt{2}-1\right)\left(2^{1-\frac{d}{2}}+1\right)\). After setting \(d=11\) and \(d=12\) into the inequality, we see that they do not satisfy it. The left-hand side is always greater than \(\frac{37}{87}\), and the right-hand side is decreasing. If \(d\geq 13\), then the right-hand side is even smaller than \(\frac{37}{87}\). We obtain a contradiction. Therefore, if \(d\geq 11\), then the global maximum of \(W_{PB}\) does not occur on \([0.74,1)\). \(W_{PB}^{\prime}=\frac{-\frac{1}{2^{2}}(k^{d}+\left(\frac{1}{2}-k\right)^{d}+1) -\left(\frac{1}{2}+1\right)\left(dk^{d-1}+d\left(-\frac{1}{2}-1\right)\left( \frac{1}{2}-k\right)^{d-1}\right)}{\left(k^{d}+\left(\frac{1}{2}-k\right)^{d} +1\right)^{2}}\) is continuous on \(\left(\frac{\sqrt{2}}{2},1\right)\). Assume, for the sake of contradiction, that if \(d\geq 11\), then \(W_{PB}^{\prime}=0\) at some \(k\in\left(\frac{\sqrt{2}}{2},0.74\right]\), we obtain:  
\[\Big{(}dk^{2}+d+k-1\Big{)}\bigg{(}\frac{1}{k}-k\bigg{)}^{d}=(1-k)\Big{(}dk^{d +1}+dk^{d}+k^{d}+1\Big{)} \tag{5}\]  
Using the endpoints of the interval in Equation (5), we obtain an upper bound of the left-hand side and a lower bound of the right-hand side. \(\left(0.74^{2}d+d+0.74-1\right)\left(\frac{1}{\frac{\sqrt{2}}{2}}-\frac{\sqrt {2}}{2}\right)^{d}\geq(1-0.74)\bigg{(}d\left(\frac{\sqrt{2}}{2}\right)^{d+1}+ d\left(\frac{\sqrt{2}}{2}\right)^{d}+\left(\frac{\sqrt{2}}{2}\right)^{d}+1 \bigg{)}\)  
gives \(\left(3219-325\sqrt{2}\right)d-650\Big{(}2^{d/2}+2\Big{)}\geq 0\). If \(d\doteq 10.95\), then the left-hand side is \(0\), if \(d\geq 7.23\), then the left-hand side is decreasing. Therefore, if \(d\geq 11\), then the left-hand side is always negative. This is a contradiction. Hence, if \(d\geq 11\), then \(W_{PB}\) has no local extremum on \(\left(\frac{\sqrt{2}}{2},0.74\right)\). Therefore, if \(d\geq 11\), then the global maximum of \(W_{PB}\) occurs at \(P\) or at \(B\), but \(G\) does not attain the global maximum at \(P\) or at \(B\). Therefore, if \(d\geq 11\), then the global maximum of \(G\) must occur on the interior of \(PC\). If \(2\leq d\leq 10\), then \(W_{PB}\) has exactly one critical point inside \(PB\). Proof. : We remove the fractions from Equation (5) by multiplying it by \(k^{d}\). The global maximum of \(G\) does not occur on \(BC\), so we remove the root \(k=1\) by dividing the equation by \((k-1)\), and we obtain:  
If \(d\) is odd, then we divide the equation additionally by \((k+1)^{2}\). If \(d=2\), then we divide the equation additionally by \((2k+1)\). Removing these roots is not necessary, but we reduce the degree of the polynomial in this way.