Htykuut

A problem from a Moscow Olympiad states : let $a,b,c$ be real numbers such that $(a+b+c)c<0$. Show that $b^2-4ac>0$. There is a well known proof of this applying the IVT on the polynomial $f(x) = ax^2+bx+c$. Is there any proof that does not involve any calculus, including IVT?

The question emerged from a reddit post
https://www.reddit.com/r/math/comments/9znumt/theorems_of_single_variable_calculus/

We have
$$
(b^2 - 4ac) + 4(a+b+c)c = (b+2c)^2 ge 0 \
implies b^2 - 4ac ge - 4(a+b+c)c > 0
$$

To give proper credit: The above approach was found after reading
guest's answer
and is just a simplification of that solution.

Here is a proof that works for any ordered commutative ring:

Let $a,b,c$ be rationals, reals or elements of any non-trivial ordered commutative ring. Assume that $(a+b+c)c<0$. We show that $b^2-4ac>0$.

Solution:

First, observe that the hypothesis implies $(a+b)c<-c ^2$ and that $cneq 0$.
Then form the matrix
$$M = left( begin{array}{cc}
2c & b \
b & 2a \
end{array}right)$$
whose determinant is precicely $4ac-b^2$, and the matrix
$$ S = left( begin{array}{cc}
1 & 0 \
2 & 1 \
end{array}right)$$
whose determinant is $1$. Now from the multiplicativity of the determinant $det(SM) =det(S)det(M)=det(M)$. But
$$det(SM) = det
left( begin{array}{cc}
2c & b \
4c+b & 2(a+b) \
end{array}right) = 4c(a+b) - b^2-4bc.$$

But recall that $c(a+b)<-c^2$ so $$4c(a+b)-b^2-4bc < -4c^2-b^2 -4bc = -(b+2c)^2.$$
The last element is always non-positive and the inequality is strict, so $det(M)=4ac-b^2<0$.

If all you know is completing the square:

Suppose $b^2-4acleq0$. Then we know the quadratic $ax^2+bx+c$ has at most one real root.

By completing the square we can rewrite $ax^2+bx+c=aleft(left(x+frac{b}{2a}right)^2+d^2right)$ for some real $d$.

We have $c=frac{b^2}{4a}+ad^2$. Returning to the original expression, complete all the squares:

$$begin{align}(a+b+c)c&=left(a+b+frac{b^2}{4a}+ad^2right)left(frac{b^2}{4a}+ad^2right)\
&=left(a+b+frac{b^2}{4a}right)frac{b^2}{4a}+left(ad^2frac{b^2}{4a}+left(a+b+frac{b^2}{4a}right)ad^2right)+left(ad^2right)^2\
&=frac{b^2}{4}left(1+frac{b}{a}+frac{b^2}{4a^2}right)+d^2left(a^2+ab+frac{b^2}{2}right)+left(ad^2right)^2\
&=frac{b^2}{4}left(1+frac{b}{2a}right)^2+d^2left(a+frac{b}{2}right)^2+left(ad^2right)^2\
&geq0text{.}
end{align}$$
Thus by contrapositive, $(a+b+c)c<0,Rightarrow, b^2-4ac>0$.

Here's a silly proof deriving a contradiction from assuming that $(a + b + c)c lt 0 $ and $b^2-4ac le 0$ are simultaneously true.

1) By assumption

$$ (a + b + c) c lt 0 $$

2) Negation of goal: $ b^2 - 4ac gt 0$

$$ b^2 - 4ac le 0 $$

3) From 2 by addition

$$ b^2 le 4ac $$

4) expand 1

$$ ac + bc + c^2 < 0 $$

5) From 4 by addition

$$ ac < -bc - c^2 $$

6) From 5 by multiplication

$$ 4ac < -4bc -4c^2 $$

7) transitivity $x le y lt z$ implies $x lt z$

$$ b^2 lt -4bc - 4c^2 $$

8) from 7 by addition.

$$ b^2 + 4bc + 4c^2 gt 0 $$

9) (7) is a perfect square

$$ (b + 2c)^2 < 0 $$

10) By assumption, $a, b, c in mathbb{R}$, so the square of $(b + 2c)$ cannot be negative.

$$ bot $$

Therefore, $b^2 - 4ac gt 0$ .