2a分之1+2b分之1+2c分之1大于等于b+c分之1+c+a分之1+a+b分之1 怎么求证啊

1/2a=5ab/10a²b
1/5a²b=2/10a²

1/2a-1/(a-b)*[(a-b)/2a-a^2+b^2]
=1/2a-1/(a-b)*[(a-b)/2a-(a^2-b^2)]
=1/2a-1/(a-b)*(a-b)/2a+1/(a-b)*(a^2-b^2)
=1/2a-1/(a-b)*(a-b)/2a+1/(a-b)*(a-b)(a+b)
=1/2a-1/2a+(a+b)
=a+b
=3-2√2+3√2-3
=√2

首先(a-b)^2≥0 =>（a+b)^2≥4ab
因此1/4a+1/4b =(a+b)/4ab ≥(a+b)/(a+b)^2
即 1/4a+1/4b≥1/(a+b) (1)
同理 1/4a+1/4c≥1/(a+c) (2)
1/4b+1/4c≥1/(b+c) (3)
(1)+(2)+(3)得
1/2a+1/2b+1/2c≥1/(b+c)+1/(c＋a)+1/(a＋b)

(1)M/（5分之M+3分之M）=M/(M/5+M/3)=M/[(8/15)M]=15/8
(2)[3/A-1/(2A)]*4A=[(3-1/2)/A]*4A=(3-1/2)*4=12-2=10

首先提一个意见：你没有说清楚这是小学、初中还是高中的题.其次,给你答案:
1.如果a=1/2或者a=-1/2,则二者相等；
2.如果a>1/2或者-1/2

根下2a+1有意义,则2a+1大于等于0,a大于等于—2分之1
同理对于5-2a有a小于2.5
所以a等于0,1,2
关键是你题貌似不完整啊...

a方程x平方加3x加1等于0的根．
a=-3±√(3²-2)/2=-3±1/2√14
(a平方减4a+4)分之a平方-4减a平方-2a分之1)除以a平方-2a分之2
=[(a-2)(a+2)/(a-2)²-1/a(a-2)]÷2/a(a-2)
=[(a+2)/(a-2)-1/a(a-2)]×a(a-2)/2
=[(a²+2a-1)/a(a-2)]×a(a-2)/2
=(a²+2a-1)/2 a=-3±1/2√14
=[(-3±1/2√14)²+2*(-3±1/2√14)-1]/2
=(9±3√14+7/2-6±√14-1)/2
=(23/2±4√14)/2
=23/4±2√14

m/（m/5+m/3)*(a/3-1/2a)*4a=1/1/5+1/3)*(1/3-1/2)*4=15/8*(-1/6)*4=-5/4

原式=1-(a-1)/a÷(a²-1)/a(a+2)
=1-(a-1)/a×a(a+2)/(a+1)(a-1)
=1-(a+2)/(a+1)
=(a+1-a-2)/(a+1)
=-1/(a+1)
令a=3
原式=-1/(3+1)=-1/4