Finding $f$ given that all roots of $x^8 - 4x^6 + 7x^6 + ax^5 + bx^4 + cx^3 + dx^2 + ex + f$ are positive real numbers

Question

Assume $a,b,c,d,e,f$ are real numbers such that all the roots of $x^8 - 4x^7 + 7x^6 + ax^5 + bx^4 + cx^3 + dx^2 + ex + f$ are positive real numbers. Find all possible values of $f.$

I'm pretty sure that Vieta's plays a vital part in here, but I'm not sure how to set it up. Can someone give me a hint please?

Well, if you know that Vieta's formulas play a role, what do they say about the roots of this polynomial ? — D. Thomine, Oct 02 '20 at 14:42
Hint: Newton's inequality look careful at the condition when the inequality achieves equality. — achille hui, Oct 02 '20 at 15:04

Will Jagy · Answer 1 · 2020-10-02T16:32:41.940

1

$$ \sum_{i=1}^8 \left( x_i^2 - x_i + \frac{1}{4} \right) = 0 $$

$$ \sum_{i=1}^8 \left( x_i - \frac{1}{2} \right)^2 = 0 $$

each summand is zero.

In general, with Vieta $$ \sigma_1 = \sum_{i = 1}^n x_i $$ and

$$ \sigma_2 = \sum_{1 \leq i < j \leq n} x_i x_j $$ we have $$ \sigma_1^2 = 2 \sigma_2 + \sum_{i=1}^n x_i^2 $$ and may thus express $\sum_{i=1}^n x_i^2 $

edited Oct 02 '20 at 16:32

answered Oct 02 '20 at 15:08

Will Jagy

139,541

I'm a bit confused as to how this will help. – questionasker Oct 02 '20 at 16:42
@questionasker sorry to hear that – Will Jagy Oct 02 '20 at 17:10

score 0 · Answer 2 · answered Oct 02 '20 at 14:52

0

Assuming that the second term is $-4x^7$

$x_1+x_2+x_3+...+x_8=4, x_1 x_2 x_3...x_8=f$ Since roots are real positive AM-GM will hold, then $$\frac {x_1+x_2+x_3+...+x_8}{8} \ge (x_1.x_2.x_3....x_8)^{1/8} \implies f \le \frac{1}{128}$$

answered Oct 02 '20 at 14:52

Z Ahmed

43,235

Finding $f$ given that all roots of $x^8 - 4x^6 + 7x^6 + ax^5 + bx^4 + cx^3 + dx^2 + ex + f$ are positive real numbers

2 Answers2