Can You Prove $ab\le 1$ Given $(x+a)(x+b)=9$?

[anemone]

Here is this week's POTW:

-----

The equation $(x+a) (x+b) = 9$ has a root $a+b$.

Prove that $ab\le 1$.

-----

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 the following members for their correct solution!(Cool)

1. Olinguito
2. castor28
3. kaliprasad

Solution from Olinguito:

Spoiler

We have
$$(a+b+a)(a+b+b)=9$$

$\implies\ 2b^2+5ba+2a^2-9\ =\ 0$.

Treating this as a quadratic equation in $b$,

$$b\ =\ \frac{-5a\pm\sqrt{(5a)^2-4(2)(2a^2-9)}}4\ =\ \frac{-5a\pm3\sqrt{a^2+8}}4$$

$\implies\ ab\ =\ \dfrac{-5a^2\pm3a\sqrt{a^2+8}}4.$

We want to show that this is less than or equal to $1$. Assume to the contrary that it is greater than $1$:

$$\dfrac{-5a^2\pm3a\sqrt{a^2+8}}4\ >\ 1$$

$\implies\ \pm3a\sqrt{a^2+8}\ >\ 4+5a^2.$

The RHS is greater than $0$ so we can square the inequality:

$$9a^2(a^2+8)\ >\ 25a^4+40a^2+16$$

$\implies\ 0\ >\ 16a^4-32a^2+16\ =\ 16(a^2-2)^2.$

We have a contradiction; therefore the assumption that $ab>1$ is false. Hence $\boxed{ab\leqslant1}$.

Alternative solution from castor28:

Spoiler

The equation can be written as $x^2+(a+b)x+ab=9$. As $x=a+b$ satisfies the equation, we have:
$$
2(a+b)^2+ab=9
$$

Using the identity $(a+b)^2 = (a-b)^2+4ab$, we get:
\begin{align*}
2(a-b)^2 + 9ab&=9\\
9ab&\le9\\
ab&\le1
\end{align*}