(1)设a,b,c为正数，求证:(bc/a)+(ca/b)+(ab/c)>=a+b+c(2)设a>b>c,求证:[4/(a-b)]+[9/(b-c)]>=[25/(a-c)]请求详解，谢谢
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1)设a,b,c为正数，求证:(bc/a)+(ca/b)+(ab/c)≥a+b+c(bc/a)+(ca/b)+(ab/c)-(a+b+c)=[(bc)^2+(ac)^2+(ab)^2-abc(a+b+c)]/abc=(1/2)[(ab-bc)^2+(ab-ac)^2+(bc-ac)^2]/abc≥0 (∵a,b,c＞0)得证(2)设abc,求证:[4/(a-b)]+[9/(b-c)]≥25/(a-c)[4/(a-b)]+[9/(b-c)]-[25/(a-c)]=[4(b-c)(a-c)+9(a-b)(a-c)-25(a-b)(a-c)]/(a-b)(b-c)(a-c)=(9a^2+25b^2+4c^2-30ab+12ac-20bc)/(a-b)(b-c)(a-c)=(3a-5b+2c)^2/(a-b)(b-c)(a-c)∵a＞b＞c∴(a-b)(b-c)(a-c)＞0∴上式≥0即[4/(a-b)]+[9/(b-c)]-[25/(a-c)]≥0∴[4/(a-b)]+[9/(b-c)]≥25/(a-c)得证。