求证：a+b+c大于等于3×三次根号abc

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...