a,b,c属于正实数,a+b+c=1,求证(1/a-1)(1/b-1)(1/c-1)>=8
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a,b,c属于正实数,a+b+c=1,求证(1/a-1)(1/b-1)(1/c-1)>=8
a,b,c属于正实数,a+b+c=1,求证(1/a-1)(1/b-1)(1/c-1)>=8
a,b,c属于正实数,a+b+c=1,求证(1/a-1)(1/b-1)(1/c-1)>=8
∵a,b,c为正实数,且a+b+c=1,
∴1/a-1=(a+b+c)/a-1=(b+c)/a≥(2√bc)/a>0;
1/b-1=(a+b+c)/b-1=(c+a)/b≥(2√ca)/b>0;
1/c-1=(a+b+c)/c-1=(a+b)/c≥(2√ab)/c>0,
三式相乘,得
(1/a-1)(1/b-1)(1/c-1)≥[(2√bc)/a] [(2√ca)/b] [(2√ab)/c]=8,
即(1/a-1)(1/b-1)(1/c-1)≥8,(当且仅当a=b=1/3时取等号).