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
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 $1$.