# Three Cube Packing for All Dimensions
## 2 Results
The equation \(\frac{\partial W_{2}}{\partial m}=0\) gives  
\[1=\frac{dm^{d-1}(k+m)}{k^{d}+m^{d}+1} \tag{1}\]  
Figure 2: The domain of the function \(G(k,m)\). Assume, for the sake of contradiction, that if \(d\geq 8\), then Equation (1) holds inside of the region \(C_{2}\). We calculate an upper bound of the right-hand side of Equation (1) using constraints \(1>k(k+m)\), \(k+m>1\), \(0.5\leq k\leq 1\), and \(0<m\leq\sqrt{2}/2\). We maximize the numerator \((k=m=\sqrt{2}/2)\) and minimize the denominator \((k=m=1/2)\): \(1\leq\frac{d(\sqrt{2}/2)^{d-1}\big{(}\sqrt{2}/2+\sqrt{2}/2\big{)}}{(1/2)^{d}+ (1/2)^{d}+1}\) and we obtain \(1\leq\frac{2^{d/2+1}d}{2^{d}+2}\). If \(d\geq 8\), then we obtain a contradiction. Therefore, if \(d\geq 8\), then \(W_{2}\) has no local extremum inside of the region C\({}_{2}\). Assume, for the sake of contradiction, that if \(2\leq d\leq 7\), then \(\frac{\partial W_{2}}{\partial k}=\frac{\partial W_{2}}{\partial m}=0\) inside of the region \(C_{2}\). Step 1: From Equation (1), it follows that  
\[m=\left(\frac{k^{d}+m^{d}+1}{d(k+m)}\right)^{\frac{1}{2-1}} \tag{2}\]  
We calculate a lower bound of the right-hand side of Equation (2) using \(1>k(k+m)\), \(k+m>1\), \(0.5\leq k\leq 1\), and \(0<m\leq\sqrt{2}/2\). We minimize the numerator \((k=m=0.5)\) and maximize the denominator \((k=m=\sqrt{2}/2)\):  
\[k=\left(\frac{(2k+m+1)\Big{(}k^{d}+m^{d}+1\Big{)}}{d(k+1)(k+m)}\right)^{\frac {1}{2-1}} \tag{3}\]  
We calculate a lower bound of the right-hand side of Equation (3) using constraints \(1>k(k+m)\), \(0.53\leq m\leq k\leq 1\), and \(m\leq\sqrt{2}/2\). We minimize the numerator \((k=m=0.53)\) and maximize the denominator \((k=m=\sqrt{2}/2)\). For \(2\leq d\leq 7\), the right-hand side of inequality \(k\geq\left(\frac{(2.053+0.53+1)\big{(}0.53^{d}+0.53^{d}+1\big{)}}{d\big{(} \sqrt{2}/2+1\big{)}\big{(}\sqrt{2}/2+\sqrt{2}/2\big{)}}\right)^{\frac{1}{2-1}}\) is successively greater than \(0.838,0.681,0.677,0.694,0.715,0.734\). Therefore, \(k\) is at least \(0.67\). Step 2: We repeat the calculations analogously to step 1. For \(2\leq d\leq 7\), the right-hand side of inequality \(m\geq\left(\frac{0.67^{d}+0.58^{d}+1}{\big{(}\sqrt{2}/2+\sqrt{2}/2\big{)}d} \right)^{\frac{1}{2-1}}\) is successively greater than \(0.612,0.585,0.609,0.639,0.666,0.690\); \(m\geq 0.58\). For \(2\leq d\leq 7\), the right-hand side of inequality \(k\geq\left(\frac{(2.067+0.58+1)\big{(}0.67^{d}+0.58^{d}+1\big{)}}{d\big{(} \sqrt{2}/2+1\big{)}\big{(}\sqrt{2}/2+\sqrt{2}/2\big{)}}\right)^{\frac{1}{2-1}}\) is successively greater than \(1.08,0.777,0.735,0.734,0.744,0.756\); \(k\geq 0.73\). Step 3: We repeat the calculations, for the last time. For \(2\leq d\leq 7\), the right-hand side of inequality \(m\geq\left(\frac{0.73^{d}+0.58^{d}+1}{\big{(}\sqrt{2}/2+\sqrt{2}/2\big{)}d} \right)^{\frac{1}{2-1}}\) is successively greater than \(0.661,0.611,0.627,0.651,0.675,0.697\); \(m\geq 0.61\). For \(2\leq d\leq 7\), the right-hand side of inequality \(k\geq\left(\frac{(2.073+0.61+1)\big{(}0.73^{d}+0.61^{d}+1\big{)}}{d\big{(} \sqrt{2}/2+1\big{)}\big{(}\sqrt{2}/2+\sqrt{2}/2\big{)}}\right)^{\frac{1}{2-1}}\) is successively greater than \(1.21,0.828,0.768,0.757,0.761,0.769\); \(k\geq 0.75\). If \(\frac{\partial W_{2}}{\partial k}=\frac{\partial W_{2}}{\partial m}=0\) and \(2\leq d\leq 7\), then \(k\geq 0.75\) and \(m\geq 0.61\), which implies \(1<0.75(0.75+0.61)\). This is a contradiction, since the region \(C_{2}\) holds \(1\geq k(k+m)\). Hence, if \(2\leq d\leq 7\), then \(W_{2}\) has no local extremum inside of the region \(C_{2}\). So, the global maximum of \(W_{2}\) must occur on the boundary of the region \(C_{2}\). **Claim 3**. : _The global maximum of \(W_{2}\) does not occur on \(AB\)._Proof. : \(AB\) is a part of a line \(m=1-k\) if \(k\in\left[\frac{1}{2},1\right]\). Substituting it into \(W_{2}\) we obtain \(W_{2}(k,1-k)=\frac{k+1}{k^{d}+(1-k)^{d}+1}\). Denote it by \(W_{AB}(k)\), \(k\in\left[\frac{1}{2},1\right]\). If \(k\in\left[\frac{1}{2},1\right]\), then \(W_{AB}<2\).