<s>[INST] 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.)
 [/INST] 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 $\sqrt{\frac{5}{8}}a=1 \implies a=\sqrt{\frac{8}{5}}$
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