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.