Question

a,b,c>0;a+b+c=1,求证(a+1/a)^3+(a+1/b)^3+(c+1/c)^3≥1000/9

Answer 1

用3次的幂平均不等式:x,y,z > 0,则(x³+y³+z³)/3 ≥ ((x+y+z)/3)³.

得到(a+1/a)³+(b+1/b)³+(c+1/c)³ ≥ (a+b+c+1/a+1/b+1/c)³/9 = (1+1/a+1/b+1/c)³/9.

再由1/a+1/b+1/c = (a+b+c)(1/a+1/b+1/c) = 3+a/b+b/a+b/c+c/b+c/a+a/c ≥ 9,

代入即得(a+1/a)³+(b+1/b)³+(c+1/c)³ ≥ 1000/9.

3次的幂平均不等式可以这样证明:

(x+y+z)³ = x³+y³+z³+3x²y+3y²z+3z²x+3xy²+3yz²+3zx²+6xyz.

由3元的均值不等式有x³+y³+z³ ≥ 3xyz,2x³+y³ ≥ 3x²y等等,对各项分别放缩即得.

当然如果能用Jensen不等式,直接由(x+1/x)³的凸性证明原题是最方便的.