First of all, according to the known number you copied, (1-B2/C2) should be greater than zero.

Secondly, the equation whose formula is greater than zero is less than zero. According to the proof you copied, when x = 0 and c are both greater than zero, the values of the formulas should be the same, but because 1-B2/C2 is greater than zero, it is impossible to have two values with the same minimum value at the same time. Because this minimum value can only be the vertex of the function.

Next, I guess the part you copied wrong should be the known number, because if all three known numbers are locked, then this function is a fixed function, and there will be no problems discussed in different situations.