a,b,c均为实数,且a+b+c=1.求证(abc)/(bc+ca+ab)
a,b,c均为实数,且a+b+c=1.求证(abc)/(bc+ca+ab)
数学人气:591 ℃时间:2020-02-16 07:15:24
优质解答
a,b,c应均为正实数,由a+b+c=1,得(abc)/(bc+ca+ab)=1/(1/a+1/b+1/c),将a+b+c=1代入得(abc)/(bc+ca+ab)=1/[1+(b+c)/a+1+(a+c)/b+1+(a+b)/c]=1/[3+(b+c)/a+(a+c)/b+(a+b)/c]=1/[3+b/a+c/a+a/b+c/b+a/c+b/c],由均值不等式得(b/a)*(a/b)>=2,(c/a)*(a/c)>=2,(c/b)*(b/c)>=2,故(abc)/(bc+ca+ab)
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