方程的根的取值范围

hbghlyj

\[a,b,c \inR^+,f(x) = {x^3} - (a^2+b^2+c^2)x - 2abc,\]求证:$f(x)$有三个实根;
设$f(x)$的三个根分别为${{{x}}_0},{x_1},{x_2},{x_0} \geqslant {x_1} \geqslant {x_2},$求证:$\frac{{{{x}}_1^3 + {{x}}_2^3}}{x_0^3}\in(-1,- \frac{1}{4}]$

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最大值证明如下
当$a=b=c$时,$\frac{{{{x}}_1^3 + x_2^3}}{{x_0^3}} = \frac{{2 \cdot {{\left( { - 1} \right)}^3}}}{{{2^3}}} =  - \frac{1}{4}$.下证$\frac{{{{x}}_1^3 + x_2^3}}{{x_0^3}} \leqslant  - \frac{1}{4}$
由韦达定理${{{x}}_0} + {{{x}}_1} + {{{x}}_2} = 0,{{{x}}_0}{x_1}{x_2} = 2abc,x_0^3 + x_1^3 + x_2^3 = 3{x_0}{x_1}{x_2} = 6abc,$
$\frac{{{{x}}_1^3 + x_2^3}}{{x_0^3}} =  - 1 + \frac{{6abc}}{{x_0^3}} \leqslant  - \frac{1}{4} \Leftrightarrow {x_0} \geqslant 2\sqrt[3]{{abc}},$
又$f(x)$在$\left[ {0,\sqrt {\frac{{{a^2} + {b^2} + {c^2}}}{3}} } \right]$上严格递减,在$\left[ {\sqrt {\frac{{{a^2} + {b^2} + {c^2}}}{3}} , + \infty } \right)$上严格递增,
故$f\left( {\sqrt {\frac{{{a^2} + {b^2} + {c^2}}}{3}} } \right) < f\left( 0 \right) < 0$，由于$\mathop {\lim }\limits_{{{x}} \to  + \infty } f\left( x \right) =  + \infty ,$只需证${{f}}\left( {2\sqrt[3]{{abc}}} \right) \leqslant 0 \Leftrightarrow 8{{abc}} \leqslant 2\sqrt[3]{{abc}}\left( {{a^2} + {b^2} + {c^2}} \right) + 2abc \Leftrightarrow 3\sqrt[3]{{{a^2}{b^2}{c^2}}} \leqslant {a^2} + {b^2} + {c^2}$,由均值不等式成立.

Aluminiumor

有下确界 $-1$.
$$
\begin{align*}
&x_0^3+x_1^3+x_2^3\\
=&3x_0x_1x_2+(x_0+x_1+x_2)(x_0^2+x_1^2+x_2^2-x_0x_1-x_1x_2-x_2x_0)\\
=&6abc>0
\end{align*}
$$
$$\Longrightarrow \frac{x_1^3+x_2^3}{x_0^3}>-1$$
取 $abc\rightarrow0,x_1\rightarrow-\infty,x_2=0,x_0\rightarrow-x_1,\frac{x_1^3+x_2^3}{x_0^3}\rightarrow-1$