已知x,y,z都是大于1的正数,m>0
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∵x,y,z为正数∴利用柯西不等式(x+2y+3z)(1/x+2/y+3/z)>=(1+2+3)²所以1/x+2/y+3/z>=(1+2+3)²/(x+2y+3z)=18所以1/X
我认为用‘柯西不等式’更为简便.对于三维形式的柯西不等式可得:(a^2+b^2+c^2)(d^2+e^2+f^2)>=(ad+be+cf)^2{1/[(XY)^(1/2)]}+{1/[(YZ)^(1/
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*[
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)
x,y,z均为正数,xy+yz+zx=1,求x+y+z的最小值设M=2(x+y+z)² 则M=2x²+2y²+2z²+4xy+4yz+4zx=(x²
∵logxm=24,logym=40,logxyzm=12,∴1logxm=24,1logym=40,1logxm+logym+logzm=12.∴1124+140+logzm=12,解得logzm=
1、设x-1=a,y-1=b,z-1=c;则x=a+1,y=b+1,z=c+1.则原式可化为(a+1)+(b+1)+(c+1)+3/a+3/b+3/c=2(根号(a+3)+根号(b+3)+根号(c+3
设Y=a则x=3a,z=2a代入3a方+2a方+6a方=99a=3x=9y=3z=6代入得507
(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.
证:x立方+y立方+z立方-3xyz=0(x+y)立方+z立方-3xy(x+y)-3xyz=0(x+y+z)[(x+y)平方-z(x+y)+z平方]-3xy(x+y+z)=0(x+y+z)(x平方+2
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
左边=xy(x+y)+yz(y+z)+zx(z+x)=1/z(x+y)+1/y(x+z)+1/x(x+y)=x/z+z/x+y/x+x/y+z/y+y/z因为x,y,z都是正数,x/z+z/x=(√x
如此简单3^x=4^y=6^z=t-->x=10g3(t)y=log4(t)z=log6(t)-->1/x=logt(3)1/y=logt(4)1/z=logt(6)-->1/z-1/x=logt(6
这是道竞赛题我在电脑前没有笔,所以无法给出正确结果,但可以给你思路设f(t)=(t-x)(t-y)(t-z)则f(t)=t^3-(x+y+z)t^2+(xy+yz+zx)t-xyz代入x+y+z=1,
由题意得log(m)x=1/24,log(m)y=1/40,log(m)xyz=1/12所以log(m)xyz-log(m)x-log(m)y=log(m)y=1/12-1/24-1/40=1/60l
x+2y=1×(x+2y)=(1/x+9/y)(x+2y)=1+9x/y+2y/x+18=19+9x/y+2y/x>=19+2√(9x/y×2y/x)=19+6√2最小值为19+6√2
×=负3y=1z=2(1)X的平方+y的平方+Z的平方-2xy-2xz-2yz=(-3)×(-3)+1×1+2×2+6+12-4=9+1+4+14=-282)(x-y)(y-z)(z-x)分之xyz=
因为x+2y=1所有乘以1当然就相等啊1/x+1/y=(x+2y)(1/x+1/y)x+2y=1所以1/x+1/y=(1/x+1/y)(x+2y)=1+2y/x+x/y+2=3+(2y/x+x/y)x
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