Question

giúp mình làm bài  27 với

 

Answer 1

\(\dfrac{1}{a+b-c}+\dfrac{1}{a-b+c}\ge\dfrac{4}{\left(a+b-c\right)+\left(a-b+c\right)}=\dfrac{2}{a}\)

Tương tự: \(\dfrac{1}{a+b-c}+\dfrac{1}{b+c-a}\ge\dfrac{2}{b}\) ; \(\dfrac{1}{a-b+c}+\dfrac{1}{b+c-a}\ge\dfrac{2}{c}\)

Cộng vế:

\(2\left(\dfrac{1}{a+b-c}+\dfrac{1}{a-b+c}+\dfrac{1}{b+c-a}\right)\ge2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

\(\Leftrightarrow...\)

Dấu "=" xảy ra khi tam giác đều