用比较法证明，若a+b+c=1,a,b,c∈R+,求证：a/bc+b/ac+c/ab+bc/a+ca/b+ab/c≧10 (1)证明 欲证(1)式，只需证下列两式成立即可a/bc+b/ac+c/ab≧9 (2-1)bc/a+ca/b+ab/c≧1 (2-2)(2-1) <==>a^2+b^2+c^2-9abc≧0 (3-1)<==>(a+b+ 展开
用比较法证明，若a+b+c=1,a,b,c∈R+,求证：a/bc+b/ac+c/ab+bc/a+ca/b+ab/c≧10 (1)证明 欲证(1)式，只需证下列两式成立即可a/bc+b/ac+c/ab≧9 (2-1)bc/a+ca/b+ab/c≧1 (2-2)(2-1) <==>a^2+b^2+c^2-9abc≧0 (3-1)<==>(a+b+c)*(a^2+b^2+c^2)-9abc≧0 (3-2)由均值不等式得:(a+b+c)*(a^2+b^2+c^2)≧3(abc)^(1/3)*3(abc)^(2/3)=9abc，(2-2) <==>b^2*c^2+c^2*a^2+a^2*b^2-abc≧0 (4-1)<==>b^2*c^2+c^2*a^2+a^2*b^2-abc(a+b+c)≧0 (4-2)<==>a^2*(b-c)^2+b^2*(c-a)^2+c^2*(a-b)^2≧0.显然上式成立. 收起
