Product of numbers in any two cells sharing a side is $2$ In a $3\times 3$ square, every cell has a positive number. The product of numbers in any two cells sharing a side is exactly $2$. What is the minimum sum of all the numbers?
We may color the cells in chessboard fashion, and write $a$ in black and $b$ in white cells, where $ab=2$. The sum of all the numbers is $5a+4b\geq 2\sqrt{20ab}=4\sqrt{10}$. Inequality holds when $5a=4b$ and $ab=2$. But how to show that it suffices to consider this configuration? (It makes sense since it's the extreme configuration, but I'm not sure how to prove it.)
 A: The point is once you fix a cell number, all the neighbours (and hence, all cells) are fixed.
Suppose then the center element is $a$. Then the four neighbours must be $b=2/a$. And then the corners must be $a$. Hence the total sum is $5a + 8/a$. You must find $a$ that minimizes this. 
You can do that by analysis, or using AM-GM: $\frac{1+x^2}{2}\ge x$ with equality at $x=1$
$$5a + 8/a=\frac{8}{a} \left(\frac{5}{8}a^2 + 1\right)=\frac{8}{a}\left(\left(\sqrt{\frac{5}{8}}a\right)^2 + 1\right)\ge  \frac{8}{a} 2 \,\sqrt{\frac{5}{8}}a= 4\sqrt{10}$$
And the minimum happens at $\sqrt{\frac{5}{8}}a=1 \implies a=\sqrt{\frac{8}{5}}$
A: Let $a_{11} = a$, and $a_{12} = b$, then the condition that adjacent cells product is $2$ gives the following $3\times 3$ matrix:
$\begin{bmatrix} a & b & a\\ b & a & b\\ a & b & a\end{bmatrix}$ with $ab = 2, a > 0, b > 0 \Rightarrow \displaystyle \sum_{i,j=1}^3 a_{ij} = 5a + 4b = 5a + \dfrac{8}{a} = f(a)$.
$f'(a) = 5 - \dfrac{8}{a^2} = 0 \iff  a^2 = \dfrac{8}{5} \iff a = \dfrac{2\sqrt{10}}{5} \Rightarrow b = \dfrac{2}{a} = \dfrac{\sqrt{10}}{2} \Rightarrow f_{\text{min}} = 5\left(\dfrac{2\sqrt{10}}{5}\right) + 4\left(\dfrac{\sqrt{10}}{2}\right) = 4\sqrt{10}$.
