Vì \(a;b;c\) là 3 cạnh của tam giác nên \(a;b;c>0\)
Ta có: \(\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}=\dfrac{a^2}{ab+ac}+\dfrac{b^2}{ab+bc}+\dfrac{c^2}{ac+bc}\)
Ta sẽ chứng minh:
\(\dfrac{a^2}{ab+ac}+\dfrac{b^2}{ab+bc}+\dfrac{c^2}{ac+bc}\ge\dfrac{\left(a+b+c\right)^2}{2\left(ab+bc+ac\right)}\ge\dfrac{3}{2}\)
Thật vậy,áp dụng bđt Cauchy Schwarz cho 3 số dương ta có:
\(\dfrac{a^2}{ab+ac}+\dfrac{b^2}{ab+bc}+\dfrac{c^2}{ac+bc}\ge\dfrac{\left(a+b+c\right)^2}{2\left(ab+bc+ac\right)}\)
Như vậy cần chứng minh: \(\dfrac{\left(a+b+c\right)^2}{2\left(ab+bc+ac\right)}\ge\dfrac{3}{2}\Leftrightarrow\dfrac{\left(a+b+c\right)^2}{ab+bc+ac}\ge3\)
\(\Rightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ac\right)\)
\(\Rightarrow a^2+b^2+c^2+2ab+2bc+2ac\ge3ab+3bc+3ac\)
\(\Rightarrow a^2+b^2+c^2\ge ab+bc+ac\) *ĐÚNG*
Dấu "=" xảy ra khi: \(a=b=c\)