已知正数x,y满足z=1 4x
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构造法:已知条件可变为1/xy+1/yz+1/xz=1要求的是1/根号(xy)+1/根号(yz)+2/根号(xz)的最大值构造1/xy+a≥2根号a*1/根号(xy)1/yz+a≥2根号a*1/根号(
配凑柯西不等式1/(x+y)+1/(y+z)+1/(z+x)≤[1/2(xy)^0.5]+[1/2(yz)^0.5]+[1/2(zx)^0.5]=(1/2){1*[z/(x+y+z)]^0.5+1*[
∵正数x、y,满足8x+1y=1,∴x+2y=(x+2y)(8x+1y)=10+xy+16yx≥10+2xy×16yx=18.当且仅当x>0,y>0,8x+1y=1,xy=16yx,解得x=12,y=
2X+4Y+3Z=92x+4y=9-3z3x-2y+5z=113x-2y=11-5z所以x=(31-13z)/8,y=(5+z)/16当0
由柯西不等式:[(y+2z)+(z+2x)+(x+2y)][x^2/(y+2z)+y^2/(z+2x)+z^2/(x+2y)]>=(x+y+z)^2=1且有(y+2z)+(z+2x)+(x+2y)=3
4^x>0,4^y>0,4^z>0所以4^x+4^y+4^z≥3(4^x*4^y*4^z)的立方根=3*[4^(x+y+z)]的立方根=3*[4^1]的立方根所以最小值=3*(4的立方根)
因为x/y+z+y/z+x+z/x+y=1所以x/y+z=1-y/z+x-z/x+y,两边同乘以x得x^2/y+z=x-xy/z+x-xz/x+y同理y^2/x+z=y-xy/z+y-yz/x+y,z
x+y-z=6y+z-x=2z+x-y=0三式相加得x+y+z=8-得2z=2z=1-得2x=6x=3-得2y=8y=4x=3y=4z=1
这么简单的题目,你们不要老是依靠答案,要自己算出答案来,就算错了,那也是你自己算出来的,就算你骗了老师,但你同事也骗了你自己
xyz=x+y+z<3z∴xy<3由于x<y,故xy=2,x=1,y=2∴z=3
(x+y)(z+y)=xz+y(x+y+z)因xyz(x+y+z)=1=xz+1/xz=(√xy-1/√xy)²+2>=2当xy=1时取得最小值取得最小值时的x,y,z并不唯一.
柯西【x^2/(y+z)+y^2/(x+z)+z^2/(x+y)】*(y+z+x+z+x+y)≥(x+y+z)^2即x^2/(y+z)+y^2/(x+z)+z^2/(x+y)≥(x+y+z)/2=(3
∵x+2y+3z=1【要将x+2y+3z用x+2y,2y+3z,3z+x表示既可以了】∴(x+2y)+(2y+3z)+(3z+x)=2(x+2y+3z)=2∴[1/(x+2y)+4/(2y+3z)+9
仔细观察:可令5x=a4y=b3z=c那么原条件即为:a+b+c=10即求证:a^2/(b+c)+b^2/(a+c)+c^2/(a+b)>=10由柯西不等式:【(b+c)+(a+c)+(a+b)】*【
证明:1/x+4/y+9/z=(x+y+z)/x+4(x+y+z)/y+9(x+y+z)/z=14+(y/x+4x/y)+(z/x+9x/z)+(4z/y+9y/z)因为x>0,y>0,z>0所以原式
x/(y+z)+y/(z+x)+z/(x+y)=1所以x/(y+z)=1-[y/(z+x)+z/(x+y)]y/(z+x)=1-[x/(y+z)+z/(x+y)]z/(x+y)=1-[x/(y+z)+
等于0.x/(y+z)=1-[y/(z+x)+z/(x+y)]y/(z+x)=1-[x/(y+z)+z/(x+y)]z/(x+y)=1-[x/(y+z)+y/(z+x)]x2/(y+z)+y2/(z+
用几何方法做
由柯西不等式可得(x+2y+2y+3z+3z+x)(1x+2y+42y+3z+93z+x)≥(1+2+3)2,∵x+2y+3z=1,∴2(1x+2y+42y+3z+93z+x)≥36,∴1x+2y+4
已知正数x.y.z满足x+y+z=1,求证:(1):(1/x-1)(1/y-1)(1/z-1)大于等于8;(2):1/x+1/y+1/z大于等于9知道手机网友你好:你要发布问题,就把问题发完整.问的题