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1g*a+b/2+1g*b+c/2+1g*c+a/2>1ga+1gb+1gc已知,a,b,c为不全等的正数1g*(a+b/2)+1g*(b+c/2)+1g*(c+a/2)>1ga+1gb+1gc
题目详情
1g*a+b/2+1g*b+c/2+1g*c+a/2>1ga+1gb+1gc
已知,a,b,c为不全等的正数
1g*(a+b/2)+1g*(b+c/2)+1g*(c+a/2)>1ga+1gb+1gc
已知,a,b,c为不全等的正数
1g*(a+b/2)+1g*(b+c/2)+1g*(c+a/2)>1ga+1gb+1gc
▼优质解答
答案和解析
1g*a+b/2+1g*b+c/2+1g*c+a/2>1ga+1gb+1gc
(a+b/2)>a
(b+c/2)>b
(c+a/2)>c
(a+b/2)*(b+c/2)*(c+a/2)>a*b*c
1g*a+b/2+1g*b+c/2+1g*c+a/2>1ga+1gb+1gc
(a+b/2)>a
(b+c/2)>b
(c+a/2)>c
(a+b/2)*(b+c/2)*(c+a/2)>a*b*c
1g*a+b/2+1g*b+c/2+1g*c+a/2>1ga+1gb+1gc
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