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.)

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 $\sqrt{\frac{5}{8}}a=1 \implies a=\sqrt{\frac{8}{5}}$
