Question

求证：a+b+c大于等于3×三次根号abc

Answer 1

令a=x³,b=y³,c=z³.

因为 x³+y³+z³-3xyz=(x+y+z)(x²+y²+z²-xy-yz-zx)

=(x+y+z)[(x-y)²+(y-z)²+(z-x)²]/2≥0,

所以 x³+y^³+z³≥3xyz,

即 a+b+c≥3(abc)^(1/3).