# Three Cube Packing for All Dimensions
## 2 Results
Therefore, the global maximum of \(G\) does not occur at \(P\). Claim 6: The global maximum of \(W_{1}\) does not occur on \(BC\). **Proof. **\(BC\) is a part of a line \(k=1\) if \(m\in(0,1)\). Substituting it into \(W_{1}\), we obtain \(W_{1}(1,m)=\frac{m+2}{m^{d}+2}\). We denote it by \(W_{BC}(m)\), \(m\in(0,1)\). We show that, for each \(m_{0}\in(0,1)\), there exists \(k_{0}\), such that point \((k_{0},m_{0})\in C_{1}\) and \(W_{BC}(m_{0})<W_{1}(k_{0},m_{0})\) holds. If \(m\in\left[\frac{\sqrt{2}}{2},1\right)\) and \(d\geq 2\), then \(\frac{m-md}{\left(2m^{d}+1\right)\left(m^{d}+2\right)}>0\) holds. It implies \(\frac{2m+1}{2m^{d}+1}>\frac{m+2}{m^{d}+2}\). Thus, \(W_{1}(m,m)>W_{BC}(m)\) on \(m\in\left[\frac{\sqrt{2}}{2},1\right)\). Therefore, the global maximum of \(W_{1}\) does not occur on \(BC\) if \(m\in\left[\frac{\sqrt{2}}{2},1\right)\) and \(d\geq 2\). If \(m\in\left(0,\frac{\sqrt{2}}{2}\right)\), then the line \(k=1-\frac{\pi}{3}\) is a part of \(C_{1}\) (see the dash-dotted line in Figure 2). Assume, for the sake of contradiction, that \(W_{1}(1-\frac{m}{3},m)\leq W_{BC}(m)\) holds at some \(m\in\left(0,\frac{\sqrt{2}}{2}\right)\). From \(\frac{2m/3+2}{m^{d}+\left(1-m/3\right)^{d}+1}\leq\frac{m+2}{m^{d}+2}\), we obtain \(\left(\frac{2m}{3}+2\right)\left(m^{d}+2\right)\leq\left(m+2\right)\left(m^{ d}+\left(1-\frac{m}{3}\right)^{d}+1\right)<\left(m+2\right)\left(m^{d}+\left(1- \frac{m}{3}\right)^{\mathbf{1}}+1\right)\) and, finally, \(m^{2}<m^{d+1}\). This is a contradiction. Therefore, if \(m\in\left(0,\frac{\sqrt{2}}{2}\right)\) and \(d\geq 2\), then the global maximum of \(W_{1}\) does not occur on \(BC\). So, the global maximum of \(W_{1}\) does not occur on \(BC\). The previous claims imply that the global maximum of \(G\) must occur on interior of \(PC\) or \(PB\). 1. If \(m\in\left(\frac{\sqrt{2}}{2},1\right)\), then \(m=k\). \(G(k,k)=\frac{2k+1}{2k^{d}+1}\). We denote it by \(W_{PC}(k)\), \(k\in\left(\frac{\sqrt{2}}{2},1\right)\). 2. If \(m\in\left(0,\frac{\sqrt{2}}{2}\right)\), then \(m=\frac{1}{k}-k\). \(G(k,\frac{1}{k}-k)=\frac{\frac{1}{k+1}}{k^{d}+\left(\frac{1}{k}-k\right)^{d}+1}\). We denote it by \(W_{PB}(k)\), \(k\in\left(\frac{\sqrt{2}}{2},1\right)\). Claim 7: If \(d\leq 10\), then the global maximum of \(G\) must occur on the interior of \(PB\). If \(d\geq 11\), then \(W_{PC}\) has exactly one critical point on \(PC\). **Proof. **\(W_{PC}^{\prime}=\frac{2\left(2k^{d}+1\right)-2d\left(2k+1\right)^{d-1}}{\left(2 k^{d}+1\right)^{2}}\) is never undefined. \(W_{PC}^{\prime}=0\) gives:  
\[2(d-1)k^{d}+dk^{d-1}=1 \tag{4}\]  
Let \(h(k)\) denote the left-hand side of Equation (4). The function \(h(k)\) is increasing and continuous if \(k>0\), therefore Equation (4) holds on \(\left(\frac{\sqrt{2}}{2},1\right)\) if and only if \(h(\sqrt{2}/2)\leq 1\) and \(h(1)\geq 1\). From the first inequality, we obtain \(\left(\sqrt{2}+2\right)d\leq 2^{d/2}+2\), which only holds for \(d\geq 11\), where \(h(1)\geq 1\) holds for \(d\geq 2\). So, if \(2\leq d\leq 10\), then \(W_{PC}\) has no local extremum inside of \(PC\). Therefore, if \(2\leq d\leq 10\), then the global maximum of \(W_{PC}\) must occur at \(P\) or at \(C\), but \(G\) does not attain the global maximum at \(P\) or \(C\). Therefore, if \(2\leq d\leq 10\), then the global maximum of \(G\) must occur on the interior of \(PB\). The function \(h(k)\) is increasing and continuous if \(k>0\); therefore, if \(d\geq 11\), then Equation (4) has exactly one solution on \(\left(\frac{\sqrt{2}}{2},1\right)\). Therefore, if \(d\geq 11\), then \(W_{PC}\) has exactly one critical point on \(PC\). Claim 8: 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.