Find all non-zero real numbers $a, b, c$ such that the following polynomial has four (not necessarily distinct) positive real roots. $$P(x) = ax^4 - 8ax^3 + bx^2 - 32cx + 16c$$
Problem
Source: 2023 NZMO - New Zealand Maths Olympiad Round 1 p8
Tags: polynomial, algebra
soryn
02.09.2023 22:59
Let: p,q,rs -the four positive roots of P. p+q+r+s=8 (1); pq+qr+rs+sp=b/a (2);. pqr+pqs+prs+qrs=32*c/a (3), pqrs=16*c/a(4); from (3),(4) =>(1/p)+(1/q)+(1/r)+(1/s)=2 (5). By AM-GM: (1/4*(p+q+r+s))^4>=pqrs and this leads to pqrs<=16. (6). Also(1/4(1/p+1/q+1/r+1/s))^4>=1/p*1/q*1/r*1/s and this leads to pqrs>=16 (7); from (6),(7)=>pqrs=16 and p=q=r=s=2; (2)=>b/a=16; (4)=>c=a then (a,b,c)=(a,16a,a)
Myddguai
04.09.2023 09:34
@above, check your solution again? I think it's wrong
Let the four roots of $P(x)$ be $w$,$x$,$y$,$z$. We have by Vieta's Formulas that:
$$w+x+y+z = 8$$$$wx+xy+yz+zw+wy+xz = \frac{b}{a}$$$$wxy+xyz+yzw+zwy = \frac{32c}{a}$$$$wxyz=\frac{16c}{a}$$Let $\alpha = \frac{16c}{a}$ and $\beta = \frac{b}{a}$.
If $w, x, y, z$ are to be positive, we must have the following:
$$(wx+xy+yz+zw+wy+xz)(wxy+xyz+yzw+zwy)-6wxyz(w+x+y+z)\ge0$$by Muirhead, and
$wx+xy+yz+zw+wy+xz \le 24$ by Smoothing, since $$wx+xy+yz+zx+wy+xz \le \frac{3(w+x+y+z)^2}{8}$$
The former gives that $2\alpha\beta-48\alpha\ge0$ so $\beta \ge 24$, and the latter gives that $\beta \le 24$, so $\beta = 24$. Equality is achieved in both inequalities when $w=x=y=z$, so we have the only solution is $P(x)=a(x-2)^4$, or when $(a,b,c)=(a,24a,a)$.
soryn
04.09.2023 20:17
Ok,is (a,24a,a)..
grupyorum
04.09.2023 22:26
Let $x_1,\dots,x_4>0$ be the roots of this polynomial. By Vieta's formulas, $x_1+\cdots+x_4=8$. Moreover, $\textstyle \sum_{i<j}x_ix_j = 32c/a$ and $x_1x_2x_3x_4 = 16c/a$. So, $\textstyle \sum_i \frac{1}{x_i}=2$. Now, we have \[ 16 = \left(\sum_i x_i\right)\left(\sum_i\frac{1}{x_i}\right)\ge 16, \]using your favourite inequality (such as AM-HM or Cauchy-Schwarz). Hence, we retain equality everywhere, so $x_i=2$ identically. Thus, $P(x)=a(x-2)^4$, yielding $(a,b,c)=(a,24a,a)$ as the desired family.