Consider the top-right $3\times 3$ table $(a_{ij})_{3\times 3}$. Note that the GM of the elements in any column is the second element, and the AM of the elements of the elements in any row is the second one. Then $$\frac{\sqrt[3]{a_{11}a_{21}a_{31}}+\sqrt[3]{a_{12}a_{22}a_{32}}+\sqrt[3]{a_{13}a_{23}a_{33}}}{3}=a_{22}=\sqrt[3]{\frac{a_{11}+a_{12}+a_{13}}{3}\cdot \frac{a_{21}+a_{22}+a_{23}}{3}\cdot \frac{a_{31}+a_{32}+a_{33}}{3}}$$But by Holder, \begin{align*}(a_{11}+a_{12}+a_{13})( a_{21}+a_{22}+a_{23} )(a_{31}+a_{32}+a_{33})&\ge (\sqrt[3]{a_{11}a_{21}a_{31}}+\sqrt[3]{a_{12}a_{22}a_{32}}+\sqrt[3]{a_{13}a_{23}a_{33}})^3\\ \iff \frac{\sqrt[3]{a_{11}a_{21}a_{31}}+\sqrt[3]{a_{12}a_{22}a_{32}}+\sqrt[3]{a_{13}a_{23}a_{33}}}{3}&\le \sqrt[3]{\frac{a_{11}+a_{12}+a_{13}}{3}\cdot \frac{a_{21}+a_{22}+a_{23}}{3}\cdot \frac{a_{31}+a_{32}+a_{33}}{3}}\end{align*}So equality holds in Holder's inequality, which means the rows are all proportional. So the ratios of the GPs in these $3$ columns are equal. By a similar reasoning on any $3\times 3$ sub-grid, all the GP ratios are equal. $\blacksquare$

Another way to solve. It's intuitive that a $3\times 3$ grid will fail. So we let the grid be $$\begin{bmatrix} \frac{a-d}x & \frac{a}y & \frac{a+d}z \\ a-d & a & a+d \\ (a-d)x & ay &(a+d)z\end{bmatrix}$$ Thus we have the following system of equations $$\frac{a-d}x + \frac{a+d}y = \frac{2a} y$$ $$(a-d)x + (a+d)z = 2ay$$ We divide by $a$ for each equation, and letting $k = \frac da$, we get that these transform into $$\frac{1-k}x + \frac{1+k}z = \frac 2y$$$$(1-k)x +(1+k)z = 2y$$ For some $1>k>0$. Now let's multiply both equations. Doing some brief rearranging gives us $$\frac{(k-1)(k+1)(x-z)^2}{xz} = 0$$ This implies that $x = z$ because $0<k<1$. Plugging this back in quickly yields $x = y = z$. This implies the problem.