x,y都大于零,若x+y=5,求(x+1/x)(y+1/y)的最小值?
来源:学生作业帮 编辑:百度作业网作业帮 分类:数学作业 时间:2024/07/23 17:35:39
x,y都大于零,若x+y=5,求(x+1/x)(y+1/y)的最小值?
![x,y都大于零,若x+y=5,求(x+1/x)(y+1/y)的最小值?](/uploads/image/z/15330702-30-2.jpg?t=x%2Cy%E9%83%BD%E5%A4%A7%E4%BA%8E%E9%9B%B6%2C%E8%8B%A5x%2By%3D5%2C%E6%B1%82%28x%2B1%2Fx%29%28y%2B1%2Fy%29%E7%9A%84%E6%9C%80%E5%B0%8F%E5%80%BC%3F)
解法一:
x、y>0,且
5=x+y≥2√(xy)→√(xy)≤5/2.
依Cauchy不等式,得
(x+1/x)(y+1/y)
≥[√(xy)+1/√(xy)]^2
=(5/2+2/5)^2
=841/100.
故所求最小值为:841/100.
解法二:
构造下凸函数f(t)=ln(t+1/t),则
f(x)+f(y)≥2f[(x+y)/2]
→ln(x+1/x)+ln(y+1/y)≥2ln[(x+y)/2+2/(x+y)]
→ln[(x+1/x)(y+1/y)]≥ln(5/2+2/5)^2
→(x+1/x)(y+1/y)≥841/100.
故所求最小值为:841/100.
x、y>0,且
5=x+y≥2√(xy)→√(xy)≤5/2.
依Cauchy不等式,得
(x+1/x)(y+1/y)
≥[√(xy)+1/√(xy)]^2
=(5/2+2/5)^2
=841/100.
故所求最小值为:841/100.
解法二:
构造下凸函数f(t)=ln(t+1/t),则
f(x)+f(y)≥2f[(x+y)/2]
→ln(x+1/x)+ln(y+1/y)≥2ln[(x+y)/2+2/(x+y)]
→ln[(x+1/x)(y+1/y)]≥ln(5/2+2/5)^2
→(x+1/x)(y+1/y)≥841/100.
故所求最小值为:841/100.
设x,y都大于零,求【x+y][1/x+25/y]的值
1/x+4/y=1,x,y都大于0,求xy的最小值
若x大于y大于0,且x+2y=3,求1/x+1/y的最小值 )
设X大于0.Y大于0,且X+2Y=1求1/X+1/Y的最小值
设x大于0,Y大于0,且1/X+9/Y=1,求X+Y的最小值.
已知x大于0,y大于0,且1/x+9/y=1,求x+y的最小值
已知x大于0,Y大于0,且1/x+9/y=2,求x+y的最小值
已知x大于0,y大于0且8/x+2/y=1,求x+y的最小值
已知X大于0,Y大于0,且1/X+9/Y=3,求X+Y的最小值
y=x^2+5/x-2(x>2),y的最小值 已知x大于0小于三分之一求函数y=x(1-3x)的最大值?
若x大于零,y大于零,2x+y=1,则x分之1+y分之1的最小值为
若实数X Y满足3X-2Y-5=0{X大于等于1小于等于3}求Y/X的最大值和最小值!