# Consider the polynomial f(x)=x^4-4ax^3+6b^2x^2-4c^3x+d^4 where a,b,c,d are positive real numbers. Prove that if f has four positive distinct roots, then a > b > c > d?

Oct 27, 2016

See below.

#### Explanation:

Supposing that ${x}_{1} , {x}_{2} , {x}_{3} , {x}_{4}$ are distinct real
roots of $f \left(x\right)$ then

$f \left(x\right) = \left(x - {x}_{1}\right) \left(x - {x}_{2}\right) \left(x - {x}_{3}\right) \left(x - {x}_{4}\right)$

equating coefficients we have

$\left\{\begin{matrix}a = \frac{1}{4} \left({x}_{1} + {x}_{2} + {x}_{3} + {x}_{4}\right) \\ b = \frac{\sqrt{{x}_{1} {x}_{2} + {x}_{1} {x}_{3} + {x}_{2} {x}_{3} + {x}_{1} {x}_{4} + {x}_{2} {x}_{4} + {x}_{3} {x}_{4}}}{\sqrt{6}} \\ c = {\left({x}_{1} {x}_{2} {x}_{3} + {x}_{1} {x}_{2} {x}_{4} + {x}_{1} {x}_{3} {x}_{4} + {x}_{2} {x}_{3} {x}_{4}\right)}^{\frac{1}{3}} / {2}^{\frac{2}{3}} \\ d = {x}_{1}^{\frac{1}{4}} {x}_{2}^{\frac{1}{4}} {x}_{3}^{\frac{1}{4}} {x}_{4}^{\frac{1}{4}}\end{matrix}\right.$

We let to the reader as an exercise, to proof that if ${x}_{1} , {x}_{2} , {x}_{3} , {x}_{4}$ are distinct,

$\frac{1}{4} \left({x}_{1} + {x}_{2} + {x}_{3} + {x}_{4}\right) > \frac{\sqrt{{x}_{1} {x}_{2} + {x}_{1} {x}_{3} + {x}_{2} {x}_{3} + {x}_{1} {x}_{4} + {x}_{2} {x}_{4} + {x}_{3} {x}_{4}}}{\sqrt{6}} > {\left({x}_{1} {x}_{2} {x}_{3} + {x}_{1} {x}_{2} {x}_{4} + {x}_{1} {x}_{3} {x}_{4} + {x}_{2} {x}_{3} {x}_{4}\right)}^{\frac{1}{3}} / {2}^{\frac{2}{3}} > {x}_{1}^{\frac{1}{4}} {x}_{2}^{\frac{1}{4}} {x}_{3}^{\frac{1}{4}} {x}_{4}^{\frac{1}{4}}$

Oct 27, 2016

As the number of positive roots is given as 4, the maximum number

of changes in signs of the coefficients has to be 4.

If either a or c or both are negative, the number of changes in sign

would become two or none So, the sufficient conditions for having

all four roots as positive is { a > 0 and c > 0 }.

This is more relaxed than the given conditions.