Method checking to get the range of the three variables under two constraints Suppose a,b,c are real numbers and $a+b+c= 6$ , and $ab+bc+ca = 9$ , also $a<b<c$ find range of a , b, c .

My method was on eliminating c we get $a^2 + b^2 -6a - 6b +ab + 9$ = 0 so making a quadratic in a in terms of b we get a = $\frac {6-b \pm √[(b-6)^2 - 4(b-3)^2 ]}{2}$ , substituing this into the inequality of $a<b$ , and solving for range of $b$ we get it to be $(3,4)$ , but answer is its range is $(1,3)$ , and $c$ is $(3,4)$ , $a$ is $(0,1)$ respectively  , may anyone tells whats going wrong here ?in my method and how to get thr ranges of $a$ and $c$ respectively from my method only .
Note : i once tried with interchanging and making quadratic in  $b, c$ i got $b=f(c)$ then solve for $f(c)<c$ it lead me value of $c$ to be in $(3,4)$ , now solving for range of $b$ would be just checking the values $f(c)$ gets in the domain of $(3,4)$ but the problem is $f(c)$ is not actually a function because it has two values for same c  so how will we solve it? Although putting the domain give the range of b to be $(1,3)$ and then $0<6-b-c =a<1$   ,but how to check for b values as f is not a function , also whats wrong with the $a =f(b)$ part which was giving wrong range for b ?

 A: We have a plane $ a + b + c = 6 $ and the hyperboloid of two sheets $ab + ac + bc = 9 $
The intersection of the two is a circle of center $(2,2,2)$ and radius $\sqrt{6}$
Hence, the vector $v = (a,b,c) $ can be described parametrically as
$ v = (2,2,2) + \sqrt{6} \left( \cos(t) \dfrac{(1, -1, 0)}{\sqrt{2}} + \sin(t) \dfrac{(1, 1, -2)}{\sqrt{6}} \right) \\= (2 + \sqrt{3} \cos(t) + \sin(t), 2 - \sqrt{3} \cos(t) + \sin(t) , 2 - 2 \sin(t) )$
Since we want $ a \lt b \lt c $, then we want
$  \sqrt{3} \cos(t)  \lt - \sqrt{3} \cos(t) $
and
$ - \sqrt{3} \cos(t) + \sin(t) \lt - 2 \sin(t) $
The first equation implies that $ t \in (\dfrac{\pi}{2}, \dfrac{3 \pi}{2}) $
And the second eqauation implies
$ 3 sin(t) \lt \sqrt{3} \cos(t) $
Since, from the first inequality, $\cos(t) \lt 0 $, then
$ \tan(t) \gt \dfrac{1}{\sqrt{3}} $
whose solution is $ t \in [ \dfrac{7\pi}{6} , \dfrac{3\pi}{2} ] $
and this is the overall range.  What remains is to find the range of the functions $a,b,c$ over this interval.
The easiest is $c = 2 - 2 \sin(t) $ , so its range is $[ 3 , 4 ]$
Then we have $ a = 2 + \sqrt{3} \cos(t) + \sin(t) = 2 + 2 \cos(t - \dfrac{\pi}{6} )$, therefore, its range is $[0, 1] $
And finally, $ b = 2 - \sqrt{3} \cos(t) + \sin(t) = 2 - 2 \cos( t + \dfrac{\pi}{6} )$, therefore, its range is $[1, 3] $
A: Let $\,p=abc\,$ then by Vieta's relations $\,a,b,c\,$ are the roots of $\,f(x)=x^3 - 6 x^2 + 9 x - p\,$. For the cubic to have real and distinct roots, its discriminant must be positive $\,\Delta = -27 (p^2 - 4 p) \gt 0\,$ $\,\iff p \in (0,4)\,$.
Writing the cubic as $\,f(x) = x(x-3)^2 - p\,$, it is easily verified that when $\,0 \lt p \lt 4\,$:

*

*$f(0) = f(3) = -p \lt 0$


*$f(1) = f(4) = 4 - p \gt 0$
It follows that $\,f(x)\,$ has at least one root in each of the intervals $\,(0,1), (1,3), (3,4)\,$ and, since it cannot have more than three roots in total, it must have exactly one root in each interval.
