The dilemma is the sign
Let a , b and c be arbitrary real numbers such that a + b + c > 0 and b 2 < 4 a c . Which of the following is right about the sign of a ?
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3 solutions
Can we prove it with Arithmetic and Geometric Mean !!
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I am not sure. I am going to think about it. But there is still another way of doing it by defining a quadratic function f ( x ) = a x 2 + b x + c . Since b 2 − 4 a c < 0 , the function does not have real zeros. So its graph is going to be completely above the x-axis or completely below the x-axis. Since f ( 1 ) = a + b + c > 0 , then the whole graph is above the x-axis and this implies that a > 0 .
Let f ( x ) = a x 2 + b x + c , the general quadratic function. The given information tells us
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f ( 1 ) = a + b + c > 0 , so some of the graph of f lies above the x-axis.
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f ( x ) = 0 has no real solutions (since the discriminant b 2 − 4 a c < 0 ), so the graph does not cross the x-axis.
Thus all of the graph of f lies above the x-axis, showing that a > 0 : the parabola is open from above.
Technically we must deal with the special case a = 0 . But this implies b 2 < 0 , which is impossible. Therefore we can rule out this situation.
Another way to deal with the special case is to refer to your quadratic function. If a = 0 , then f ( x ) = b x + c , which is a linear function, meaning that part of the graph would lie below the x-axis, not fulfilling your first point.
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You'd have to consider the extra-special case a = b = 0 , which would allow the entire graph to lie above the x-axis :)
a=0 would have meant square of b less than 0 (not possible) Then I used quadratic form
Since b 2 < 4 a c then a = 0 . Now, taking into consideration that a + b + c = a ( 1 + a b + a c ) = a ( 1 + a b + 4 a 2 b 2 + a c − 4 a 2 b 2 ) = a ( ( 1 + 2 a b ) 2 + 4 a 2 4 a c − b 2 ) > 0 and ( 1 + 2 a b ) 2 + 4 a 2 4 a c − b 2 > 0 we obtain that a > 0 .