Roots of a Cubic Polynomial: Proving Coefficient Inequalities

[rock.freak667]

Homework Statement

In the equation $$x^3+ax^2+bx+c=0$$
the coefficients a,b and c are all real. It is given that all the roots are real and greater than 1.
(i) Prove that $$a<-3$$
(ii)By considering the sum of the squares of the roots,prove that $$a^2>2b+3$$
(iii)By considering the sum of the cubes of the roots,prove that $$a^3<-9b-3c-3$$

Homework Equations

If the roots are A,B and C then A+B+C = a/1=a
ABC= -c/a
AB+AC+BC= b/a

The Attempt at a Solution

I do not know if there are any other formula for the squares/cubes of roots other than the ones i stated above; If there are any simpler ones please tell me.
I got out parts (ii) by taking (A+B+C)=a and appropriately squaring it, but I was unable to get out parts (i) and (iii), could someone please help me prove it..thanks

 

[ZioX]

Since the roots are real we know the polynomial factors into $(x-r_1)(x-r_2)(x-r_3)$. Look at how $(x-r_1)(x-r_2)(x-r_3)$ multiplies out and look at a b and c in terms of the roots. For example, we know that c must be negative as $-r_1r_2r_3=c<0$. We actually know $c<-1$ as each of these roots are greater than 1.

 

[rock.freak667]

thanks, I will re-try it and see now

 

[dextercioby]

i) is trivial since A+B+C=-a (see the minus sign, for more see http://en.wikipedia.org/wiki/Vieta_relations)

ii)is doable as you said.

 

[dextercioby]

The point iii) is really tricky.

$$(A+B+C)^3 = A^3 +B^3 +C^3 -3ABC +3(A+B+C)(AB+AC+BC)$$

which means

$$-a^3 =A^3 +B^3 +C^3 +3c -3ab > 3+3c+9b$$ ,

where i used the fact that the sum of the cubes is larger than 3 and the fact that a is smaller than -3.

Multiply by -1 and you're done.