Question

设a,b,c都是正数且a b c=1,求证(a 1/a)²+(b+1/b)²+(c+1/c)²≧

设a,b,c都是正数且a b c=1,求证(a 1/a)²+(b+1/b)²+(c+1/c)²≧100/3

Answer 1

证法一：
若正数a、b、c满足a+b+c＝1,
则构造下凸函数f(x)＝(x+1/x)²,则
依Jensen不等式得
f(a)+f(b)+f(c)≥3f[(a+b+c)/3]＝3f(1/3)
→(a+1/a)²+(b+1/b)²+(c+1/c)²
≥3[(a+b+c)/3+3/(a+b+c)]²
＝3×(3+1/3)²
＝100/3
故原不等式得证.

证法二：
若正数a、b、c满足a+b+c＝1,
则依Cauchy不等式得
(a+1/a)²+(b+1/b)²+(c+1/c)²
≥[(a+1/a)+(b+1/b)+(c+1/c)]²/3
＝[(a+b+c)+(1/a+1/b+1/c)]²/3
＝[(a+b+c)+(a+b+c)(1/a+1/b+1/c)]²/3
≥[1+(1+1+1)²]²/3
＝100/3.
故原不等式得证。