✔ 最佳答案
題目應還有一前提,即a、b、c都是正實數。
(1)
若 t>0,則 2t^3 + 1 ≥ 3t^2
proof:由算幾不等式,(t^3+t^3+1)/3 ≥ t^2 ,得證,等號成立t=1
現令t=(abc)^(1/6)帶入,得到2(abc)^(1/2)+1 ≥ 3(abc)^(1/3)
(2)
令hint中x=(bc/a)^(1/3)、y=(ac/b)^(1/3)、z=(ab/c)^(1/3),可得
(bc/a)+(ac/b)+(ab/c)+3(abc)^(1/3) ≥
[(b/c)^(1/3)+(c/b)^(1/3)]a +[(c/a)^(1/3)+(a/c)^(1/3)]b +[(b/a)^(1/3)+(a/b)^(1/3)]c
≥ 2a+2b+2c ,等號成立a=b=c
[ 算幾不等式:(b/c)^(1/3)+(c/b)^(1/3)≥2 ]
(3)
由(1)、(2),
(bc/a)+(ac/b)+(ab/c)+2(abc)^(1/2)+1 ≥ (bc/a)+(ac/b)+(ab/c)+3(abc)^(1/3)
≥ 2(a+b+c),等號成立a=b=c=1
註:順便證明一下hint
因為hint中是對稱式,不妨令x≥y≥z
那麼(x^3+y^3+z^3+3xyz) - (yx^2+zx^2+zy^2+xy^2+xz^2+yz^2)
= (x^3+xyz-yx^2-zx^2)+(y^3+xyz-zy^2-xy^2)+(z^3+xyz-xz^2-yz^2)
= x(x-y)(x-z) + y(y-z)(y-x) + z(z-x)(z-y)
= (x-y) [ x^2-xz-y^2+yz] + z(x-z)(y-z)
= (x-y)(x-y)(x+y-z) + z(x-z)(y-z) ≥ 0