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
1个回答
展开全部
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<=-u,或a>=1-u,
∴a<=-1/4,或a>=1,为所求.
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<=-u,或a>=1-u,
∴a<=-1/4,或a>=1,为所求.
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询