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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.