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

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The equation (x+a)(x+b)=9 implies that the roots are related to the sum and product of a and b. Participants discuss proving that ab≤1 based on this relationship. Olinguito and castor28 provide correct solutions, highlighting different approaches to the proof. The discussion emphasizes the importance of understanding the implications of the roots in polynomial equations. The proof ultimately confirms the inequality ab≤1.
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Here is this week's POTW:

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

Prove that $ab\le 1$.

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Congratulations to the following members for their correct solution!(Cool)

1. Olinguito
2. castor28
3. kaliprasad

Solution from Olinguito:
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:
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*}
 
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