不等式への招待　第７章

>>615
let a+b+c=s, then
(a+b)(b+c)(c+a) ≧ (8/9)s(ab+bc+ca) ≧ (4s/3)^(3/2)･√(abc),

Left:
(a+b)(b+c)(c+a) - (8/9)s(ab+bc+ca)
= (1/9){(a+b+c)(ab+bc+ca) - 9abc}
= (1/18){a(b-c)^2 + b(c-a)^2 + c(a-b)^2}
≧0

Right:
(ab+bc+ca)^2 - 3s(abc)
= (ab+bc+ca)^2 - 3(ab･bc + bc･ca + ca･ab)
= (1/2){aa(b-c)^2 + bb(c-a)^2 + cc(a-b)^2}
≧0,

∴k=(4/3)^(3/2).