How Can You Prove a Quadratic with No Real Roots Satisfies the Inequality?

[anemone]

Here is this week's POTW:

-----

Given that the quadratic equation $(b+c)x^2+(a+c)x+a+b=0$ has no real roots. Prove that

$3a(a+b+c) \ge 4ac-b^2$

-----

Remember to read the https://mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to https://mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

Congratulations to castor28 for his correct answer, which you can find below:

Spoiler

We must prove that, if
$$
(a+c)^2 - 4(b+c)(a+b) < 0
$$
then
$$
4ac-b^2 -3a(a+b+c) \le 0
$$
We note first that, if $b=0$, the first relation reduces to:
$$ a^2-2ac+c^2 = (a-c)^2 < 0$$
which is impossible; therefore $b\ne0$. Writing $a = bx$, $c=by$ and dividing by $b^2$, we obtain the inequalities:
\begin{align*}
f(x,y)&=x^2 - 2xy - 4x + y^2 - 4y - 4 < 0\\
g(x,y) &= x*y - 3*x - 3*x^2 - 1\le0
\end{align*}
and we must prove that $f(x,y)<0$ implies $g(x,y)\le0$. To that effect, we try to find a positive (or negative) definite linear combination of $f$ and $g$.

Expanding $f + kg$, we find:
$$
f(x,y) + kg(x,y) = (- 3k + 1)x^2 + (k - 2)xy + y^2 + (- 3k - 4)x - 4y - (k + 4)
$$
Looking at the second degree terms, we see that a necessary condition for the expression to be a square is:
$$
(k-2)^2 -4(1-3k) = k^2 + 8k = 0
$$
giving $k=0$ or $k=-8$. $k=0$ simply gives $f(x,y)$, which is not a square. On the other hand, with $k=-8$, we find:
\begin{align*}
f(x,y) - 8g(x,y) &= 25x^2 - 10xy + 20x + y^2 - 4y + 4\\
&= (5x - y + 2)^2\\
&\ge0
\end{align*}
This can be written as $8g(x,y) = f(x,y)-(5x - y + 2)^2$, and shows that $f(x,y)<0$ indeed implies $g(x,y)<0$; this is slightly stronger that what is required, since the inequality is strict.

There is a geometric interpretation of this: in the graph below, the first relation corresponds to the interior of the green parabola, and the second relation corresponds to the exterior of the red hyperbola. The statement expresses the fact that the former is included in the latter.

View attachment 8254

Alternate solution of other:

Spoiler

Adding $ax^2+bx+c$ into $(b+c)x^2+(a+c)x+a+b=0$, we get

$(a+b+c)(x^2+x+1)=ax^2+bx+c$

Let $f(x)=(a+b+c)(x^2+x+1)$ and $g(x)=ax^2+bx+c$ and thus $f(x)=g(x)$. Since both $f$ and $g$ are quadratic functions that will never intersect, we know we either have $f(x)>g(x)$ or $f(x)<g(x)$ for all $x$.

WLOG, assume that $a+b+c>0$. We then have $f(1)=3(a+b+c),\,g(1)=a+b+c$, which implies $f(x)>g(x)$ for all $x$. It follows that the minimum of $f$ is greater than the minimum of $g$ where $f_{\text{minimum}}=\dfrac{3}{4}(a+b+c)$.

If $a$ is positive, then $g_{\text{minimum}}=\dfrac{4ac-b^2}{4a}$ and so $\dfrac{3}{4}(a+b+c)>\dfrac{4ac-b^2}{4a}$, i.e. $3a(a+b+c)>4ac-b^2$.

If $a$ is negative, then $f_{\text{maximum}}\ge f(1)$ and we get $\dfrac{4ac-b^2}{4a}\ge a+b+c\ge \dfrac{3}{4}(a+b+c)$, i.e. $3a(a+b+c)>4ac-b^2$.

And the proof is then followed.