Question

当a>0,b>0时,求证:a+b+1/ab大于或等于3 如何证明?

Answer 1

因为x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx) =(x+y+z)[(x-y)^2+(y-z)62+(z-x)^2]/2>=0,所以 x^3+y^3+z^3>=3xyz,对于 a + b + 1/(ab) ；且 a >0,b > 0 设 x = a^(1/3),y = b^(1/3),z = (1/ab)^(1/3) 则 利用 x^3 + y^3 + z^3 ≥ 3xyz 得到 a + b + 1/(ab) ≥ 3 * a^(1/3) * b^(1/3) * [1/(ab)]^(1/3) = 3