第1个回答 2022-05-21
证明:
1/[a^3(b+c)]=(bc)^3/(b+c),
(bc)^3/(b+c)+1/4(b+c)/(bc)≥bc(均值不等式)
(bc)^3/(b+c)≥bc-1/4(b+c)/(bc)=bc-1/4(1/c+1/b)=1/4(4bc-ab-ac),即
1/[a^3(b+c)]≥1/4(4bc-ab-ac),同理
1/[b^3(a+c)]≥1/4(4ac-bc-ab),
1/[c^3(a+b)]≥1/4(4ab-ac-bc),
上述三式相加,
1/[a^3(b+c)]+1/[b^3(a+c)]+1/[c^3(a+b)]
≥1/2(ab+bc+ca)≥1/2*3*(abc)^(2/3)=3/2,故命题得证.