由题设及基本不等式x+y+z≥3(xyz)^(1/3),可得
ab+[k/(2a)]+[k/(2b)]≥3(k²/4)^(1/3)
bc+[k/(2b)]+[k/(2c)]≥3(k²/4)^(1/3)
ca+[k/(2c)]+[k/(2a)]≥3(k²/4)^(1/3)
上面三式相加,可得
ab+bc+ca+[(1/a)+(1/b)+(1/c)]≥9(k²/4)^(1/3)
∴由题设可知,此时必有 9(k²/4)^(1/3)≥9
∴必有k²≥4
∴k≥2
∴k最小取2.