Question

已知a,b,c属于正实数,且a+b+c=1.求证：ab+bc+ca

Answer 1

证：

由均值不等式得

a²+b²≥2ab,b²+c²≥2bc,c²+a²≥2ca

(a²+b²)+(b²+c²)+(c²+a²)≥2ab+2bc+2ca

2(a²+b²+c²)≥2(ab+bc+ca)

a²+b²+c²≥ab+bc+ca

a+b+c=1

(a+b+c)²

=a²+b²+c²+2ab+2bc+2ca

≥ab+bc+ca+2ab+2bc+2ca

=3(ab+bc+ca)

(a+b+c)²=1

3(ab+bc+ca)≤1

ab+bc+ca≤1/3