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Let $x^2 + 2ax + b = 0$ and $x^2 + 2bx + a = 0$ have real roots $(a,b > 0)$, then minimum possible integral value of ab is___________

My approach is as follow

$T(x)=x^2 + 2ax + b = 0$, hence $4a^2-4b\ge 0$

$U(x)=x^2 + 2bx + a = 0$, hence $4b^2-4a\ge 0$

How do we approach from here

Samar Imam Zaidi
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    Alternative approach to the answer of b00n heT. Since $a,b > 0$, you have that $(ab) > 0 \implies (ab) \geq 1$, since the question calls for an integral value for $(ab)$. So, all that is necessary is to manually check whether $(a=1,b=1)$ satisfies the constraints. – user2661923 Sep 11 '21 at 10:32
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    In fact $a \leq b^2, b \leq a^2$ and multiplying LHS and RHS, $ab \leq a^2b^2 \implies ab \geq 1$ as $ab \ne 0$ – Math Lover Sep 11 '21 at 10:45
  • I'm sorry, but not simplifying $4a^2-4b \ge 0$ to $a^2>b$, or at least to $a^2-b \ge 0$ makes you look lazy. – jjagmath Sep 11 '21 at 12:22

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Notice that the two inequalities can be rewritten as $$\begin{cases}a^2\geq b\\ b^2\geq a \end{cases}$$ and since both $a,b>0$ and $\sqrt{\cdot}$ is increasing we can take roots and obtain the inequality $$a^2\geq b\geq \sqrt{a}$$ Now $$a^2\geq \sqrt{a}\iff a\geq 1,$$ ($a=0$ is excluded by assumtion) and by a simmetric argument for $b$ we get $b\geq 1$.

In conclusion the minimal value for $a\cdot b$ is for $a=1=b$ and is $1$.

b00n heT
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  • $\sqrt{\cdot}$ is not always increasing – MAS Sep 11 '21 at 10:37
  • What do you mean? Maybe my notation is misleading: I am saying that you can apply the square root to an inequality as it is an increasing function, I am not saying that $\sqrt{f(x)}$ is increasing in general. – b00n heT Sep 11 '21 at 10:41
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    sorry, it is my misunderstanding – MAS Sep 11 '21 at 10:51