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#?

Supposing that #x_1,x_2,x_3,x_4# are distinct real
roots of #f(x)# then

#f(x)=(x-x_1)(x-x_2)(x-x_3)(x-x_4)#

equating coefficients we have

#{(a = 1/4 (x_1 + x_2 + x_3 + x_4)),(b = 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 = (x_1 x_2 x_3 + x_1 x_2 x_4 + x_1 x_3 x_4 + x_2 x_3 x_4)^(1/3)/2^(2/3)),(d = x_1^(1/4) x_2^(1/4) x_3^(1/4) x_4^(1/4)):}#

We let to the reader as an exercise, to proof that if #x_1,x_2,x_3,x_4# are distinct,

# 1/4 (x_1 + x_2 + x_3 + x_4) > 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]> (x_1 x_2 x_3 + x_1 x_2 x_4 + x_1 x_3 x_4 + x_2 x_3 x_4)^(1/3)/2^(2/3) > x_1^(1/4) x_2^(1/4) x_3^(1/4) x_4^(1/4)#

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.