The first two formulas are treated as follows: take A as X, B as Y and C as constants, and draw the feasible range, then you can find the first range of b/a, which I have found, that is, (0,7).

The trouble of the third formula: after moving the term, divide both sides by a at the same time, and you can get b/a >; = (c/a) e (a/c), and then exchange elements so that a/c = x and b/a = y, then y >；; = (1/x) e x, and then take the derivative. It is found that when x= 1, the derivative is 0, so when X= 1, y has the minimum value e, so b/a >; =e (if you want to ask why C is uncertain as long as it is greater than the minimum value, you can have qualified C as long as it is greater than the minimum value).

To sum up: b/a belongs to

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