Question

Cho a,b,c ∈ R và a,b,c > 0

CM: \(\)\(\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} ≥ \dfrac{2}{a+b} + \dfrac{2}{b+c} + \dfrac{2}{c+a}\)

Answer 1

Ta co BDT :\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)\(\Leftrightarrow\dfrac{\left(a-b\right)^2}{ab\left(a+b\right)}\ge0\forall a,b\in R^+\)

Tuong tu cho 2 BDT con lai ta cung co:

\(\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{4}{b+c};\dfrac{1}{c}+\dfrac{1}{a}\ge\dfrac{4}{a+c}\)

Cong theo ve 3 BDT tren ta co

\(VT=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{2}{a+b}+\dfrac{2}{b+c}+\dfrac{2}{c+a}=VP\)

Dau "=" xay ra khi \(a=b=c\)