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If $(a+c)(a+b+c)<0,$ prove $$(b-c)^2>4a(a+b+c)$$

I will use the constructor method that want to know can not directly prove it?

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What is the constructor method? – Aryabhata Apr 7 '12 at 4:51
The same as your proof , construct a quadratic function。I hope that the constructor, the direct proof. Thank you – tianzhidaosunyouyu Apr 7 '12 at 4:57
up vote 6 down vote accepted

Consider the quadratic

$$ f(x) = ax^2 - (b-c)x + (a+b+c) $$

$$f(1)f(0) = 2(a+c)(a+b+c) \lt 0$$

Thus if $a \neq 0$, then this has a real root in $(0,1)$ and so

$$(b-c)^2 \ge 4a(a+b+c)$$

If $(b-c)^2 = 4a(a+b+c)$, then we have a double root in $(0,1)$ in which case, $f(0)$ and $f(1)$ will have the same sign.

Thus $$(b-c)^2 \gt 4a(a+b+c)$$

If $a = 0$, then $c(b+c) \lt 0$, and so we cannot have $b=c$ and thus $(b-c)^2 \gt 0 = 4a(a+b+c)$

And if you want a more "direct" approach, we show that $(p+q+r)r \lt 0 \implies q^2 \gt 4pr$ using the following identity:

$$(p+q+r)r = \left(p\left(1 + \frac{q}{2p}\right)\frac{q}{2p}\right)^2 + \left(r - \frac{q^2}{4p}\right)^2 + p\left(r - \frac{q^2}{4p}\right)\left(\left(1 + \frac{q}{2p}\right)^2 + \left(\frac{q}{2p}\right)^2\right)$$

If $(p+q+r)r \lt 0$, then we must have have $p\left(r - \frac{q^2}{4p}\right) \lt 0$, as all the other terms on the right side are non-negative.

Of course, this was gotten by completing the square in $px^2 + qx + r$ and setting $x=0$ and $x=1$ and multiplying.

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