Coefficients of Polynomial (Tomato subjective 69)

Problem: Suppose that the three equations \({\displaystyle{ax^2}} \) – \({\displaystyle {2bx + c = 0}} \) , \({\displaystyle{bx^2}} \) – \({\displaystyle {2cx + a = 0}} \) and \({\displaystyle{cx^2}} \) – \({\displaystyle {2ax + b = 0}} \) all have only positive roots. Show that a =b = c.
Solution: If possible let a, b, c are not all equal. \({\displaystyle{ax^2}} \) – \({\displaystyle {2bx + c = 0}} \) , \({\displaystyle{bx^2}} \) – \({\displaystyle {2cx + a = 0}} \) , \({\displaystyle{cx^2}} \) – \({\displaystyle {2ax + b = 0}} \) all have only positive roots.
So all of a, b, c cannot be its same sign as discriminant > 0 for three equations ( \({\displaystyle{b^2}} \) > \({\displaystyle {ac}} \), \({\displaystyle{a^2}} \) > \({\displaystyle {bc}} \), \({\displaystyle{c^2}} \) > \({\displaystyle {ab}} \) ).
WLOG we can assume a > b > c or a > c > b as the equations are cyclic.
Now we know,
\({\displaystyle{\frac{b\pm{\sqrt{b^2-ac}}}{a}}} \) > 0, \({\displaystyle{\frac{a\pm{\sqrt{a^2-bc}}}{c}}} \) > 0, \({\displaystyle{\frac{c\pm{\sqrt{c^2-ab}}}{b}}} \) > 0
Now there 2 possibilities either b, c both are positive or one of b, c is positive.
If b, c both are positive then, \({\displaystyle{\frac{b-{\sqrt{b^2-ac}}}{a}}} \) < 0 [ not possible ]

If one of b, c is positive then, \({\displaystyle{\frac{c-{\sqrt{c^2-ab}}}{b}}} \) < 0 [ not possible ]
So a, b, c have to be equal.

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