MT【321】分类线性规划

若二次函数$f(x)=ax^2+bx+c(a,b,c>0)$有零点,则$\min\{\dfrac{b+c}{a},\dfrac{c+a}{b},\dfrac{a+b}{c}\}$ 的最大值为____

由题意$b^2\ge 4ac,$由$a,c$的对称性只需考虑$b=max\{a,b,c\}\vee a=\max\{a,b,c\}$.
当$b=max\{a,b,c\}$时$\min\{\dfrac{b+c}{a},\dfrac{c+a}{b},\dfrac{a+b}{c}\}=\dfrac{c+a}{a}$
设$\dfrac{a}{b}=x,\dfrac{c}{b}=y$故
\begin{equation}
\left\{ \begin{aligned}
0<x& \le1\\
0<y&\le1\\
0<y&\le\dfrac{1}{4x}\\
\end{aligned} \right.
\end{equation}
由线性规划可知$z=x+y$在$(1,\dfrac{1}{4})$和$(\dfrac{1}{4},1)$处同时取最大值$\dfrac{5}{4}$.
当$a=\max\{a,b,c\}$时$\min\{\dfrac{b+c}{a},\dfrac{c+a}{b},\dfrac{a+b}{c}\}=\dfrac{c+b}{a}$
设$\dfrac{b}{a}=x,\dfrac{c}{a}=y$故
\begin{equation}
\left\{ \begin{aligned}
0<x& \le1\\
0<y&\le1\\
y&\le x\\
0<y&\le\dfrac{x^2}{4}\\
\end{aligned} \right.
\end{equation}
由线性规划可知$z=x+y$在$(1,\dfrac{1}{4})$时取最大值$\dfrac{5}{4}$.