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$b^2ge 4ac,$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
egin{equation}
left{ egin{aligned}
0<x& le1\
0<y&le1\
0<y&ledfrac{1}{4x}\
end{aligned}
ight.
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
egin{equation}
left{ egin{aligned}
0<x& le1\
0<y&le1\
y&le x\
0<y&ledfrac{x^2}{4}\
end{aligned}
ight.
end{equation}
由线性规划可知$z=x+y$在$(1,dfrac{1}{4})$时取最大值$dfrac{5}{4}$.