求证多项式有$\ge-1$的实根

hbghlyj

任意正数$a,b,c$，求证：以下四次多项式有$\ge-1$的实根$l$。
\begin{multline*}l^4 \left(a^4 b^2-2 a^2 b^4-2 a^2 b^2 c^2+b^6+b^4 c^2+b^2 c^4\right)\\+l \left(-a^4 b^2-a^4 c^2+2 a^2 b^4-2 a^2 b^2 c^2-b^6+3 b^4 c^2-3 b^2 c^4+c^6\right)\\+l^3 \left(-a^6+2 a^4 b^2+3 a^4 c^2-a^2 b^4-2 a^2 b^2 c^2-3 a^2 c^4-b^4 c^2+c^6\right)\\+l^2 \left(-2 a^6+2 a^4 b^2+2 a^4 c^2+2 a^2 b^4-2 a^2 b^2 c^2+2 a^2 c^4-2 b^6+2 b^4 c^2+2 b^2 c^4-2 c^6\right)\\+a^6-2 a^4 b^2+a^4 c^2+a^2 b^4-2 a^2 b^2 c^2+a^2 c^4\end{multline*}
也发到了MSE问问。

hbghlyj

用Mathematica验证很快

1. Resolve[ForAll[{a,b,c},a>0&&b>0&&c>0,Exists[l,l>=-1&&0==(a^4 b^2-2 a^2 b^4+b^6-2 a^2 b^2 c^2+b^4 c^2+b^2 c^4) l^4+(-a^6+2 a^4 b^2-a^2 b^4+3 a^4 c^2-2 a^2 b^2 c^2-b^4 c^2-3 a^2 c^4+c^6) l^3+(-2 a^6+2 a^4 b^2+2 a^2 b^4-2 b^6+2 a^4 c^2-2 a^2 b^2 c^2+2 b^4 c^2+2 a^2 c^4+2 b^2 c^4-2 c^6) l^2+(-a^4 b^2+2 a^2 b^4-b^6-a^4 c^2-2 a^2 b^2 c^2+3 b^4 c^2-3 b^2 c^4+c^6) l+a^6-2 a^4 b^2+a^2 b^4+a^4 c^2-2 a^2 b^2 c^2+a^2 c^4]],Reals]

复制代码

但不知怎么证明？

hbghlyj

@River Li的证明：

Sketch of a proof.

If $x = 0$, we have $Q(l) = y^2(l^2+y^2-2l+1)(l^2-3y^2+2l+1)$, and the statement is true. In what follows, assume that $x \ne 0$. We split into three cases.

• Case 1. $x > 0$ and $y^2 < 3x^2$

We have $Q(\infty) = -\infty$.

Also, we have $$Q\left(x - 1 + \frac{y^2}{x}\right) = \frac{y^4(x^2 + y^2)(x^2 + y^2 - 2x)^2}{x^4} \ge 0.$$

• Case 2. $x > 0$ and $y^2 \ge 3x^2$

We have $Q(-1) < 0$, and $Q(\infty) = \infty$.

• Case 3. $x < 0$

We have $Q(-1) < 0$.

(i) If $y^2 > 3x^2$, we have $Q(\infty) = \infty$.

(ii) If $y^2 \le 3x^2$, we have $$Q(-2x + 2) = \left[ 3y^4+(18x^2-22x+3)y^2+15x^4-30x^3+15x^2 \right] $$ $$\times \left( 3\,{x}^{2}-{y}^{2}-6\,x+3 \right) \ge 0. $$

We are done.

hbghlyj

很有技巧性啊，不懂如何想到取$x - 1 + \frac{y^2}{x}$和$-2x+2$这两点？