Question

已知函数f(x)=a╱x-x,对任意x属于(0,1),不等式f(x)f(1-x)≥1求a的取值

Answer 1

f(x)=a╱x-x,f(1-x)=a/(1-x)-(1-x),

对任意x属于(0,1),不等式f(x)f(1-x)≥1,

(a-x^2)[a-(1-x)^2]>=x(1-x),

a^2-a[x^2+(1-x)^2]+[x(1-x)]^2>=x(1-x),①

设u=x(1-x)∈(0,1/4],①变为a^2-a(1-2u)+u^2-u>=0,

即(a+u)(a+u-1)>=0,

∴a=1-u,

∴a=1,为所求.