if you are good at math,please…1.a,b,c都为正数,满足c/（a+b）＜a/（b+c）＜b/（c+a）,求a,b,c的大小关系.2.实数a,b,c满足条件a+b+c=9,求代数式（a-b）+（c-b）+（c-a）的最大值

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if you are good at math,please…1.a,b,c都为正数,满足c/（a+b）＜a/（b+c）＜b/（c+a）,求a,b,c的大小关系.2.实数a,b,c满足条件a+b+c=9,求代数式（a-b）+（c-b）+（c-a）的最大值
if you are good at math,please…
1.a,b,c都为正数,满足c/（a+b）＜a/（b+c）＜b/（c+a）,求a,b,c的大小关系.2.实数a,b,c满足条件a+b+c=9,求代数式（a-b）+（c-b）+（c-a）的最大值

if you are good at math,please…1.a,b,c都为正数,满足c/（a+b）＜a/（b+c）＜b/（c+a）,求a,b,c的大小关系.2.实数a,b,c满足条件a+b+c=9,求代数式（a-b）+（c-b）+（c-a）的最大值
c/(a+b)c(b+c)a^2-c^2+ab-bc>0-->(a-c)(a+b+c)>0-->a>c a/(b+c)a(a+c)b^2-a^2+bc-ac>0-->(b-a)(a+b+c)>0-->b>a b>a>c (a-b)^2+(c-b)^2+(c-a)^2=2(a^2+b^2+c^2-ab-bc-ca)=2(9-ab-bc-ca) 9=a^2+b^2+c^2>=-(ab+bc+ca)-->-(ab+bc+ca)

1因为c/（a+b）＜a/（b+c）＜b/（c+a） a，b，c都为正数所以c/（a+b）+1＜a/（b+c）+1＜b/（c+a）+1 所以（a+b+c）/（a+b）＜（a+b+c）/（b+c）＜（a+b+c）/（c+a）所以1/（a+b）＜1/（b+c）＜1/（c+a）所以a＞c＞b 2 因为（a+b）≥0 所以-2ab≤a+b （a-b）=a+b-2ab≤2（a+b）所以（a-b）+（...

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