Question

Cho a,b,c > 0 thỏa mãn \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\ge\frac{1}{2}\) . CMR:

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\)

Answer 1

Đề bài sai, phản ví dụ: \(a=b=c=\frac{1}{2}\Rightarrow\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=3>\frac{1}{2}\) (t/m)

Nhưng \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=6\ne1\)

Chắc người ta yêu cầu chứng minh \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge1\)

Ta có: \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\) ; \(\frac{1}{b}+\frac{1}{c}\ge\frac{4}{b+c}\) ; \(\frac{1}{a}+\frac{1}{c}\ge\frac{4}{a+c}\)

Cộng vế với vế:

\(2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge4\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\ge4.\frac{1}{2}=2\)

\(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge1\)

Dấu "=" xảy ra khi \(a=b=c=3\)