\(\dfrac{2a}{b+c}+\dfrac{2b}{c+a}+\dfrac{2c}{a+b}+\dfrac{a^2+b^2+c^2}{ab+bc+ca}\\ =2\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)+\dfrac{a^2+b^2+c^2}{ab+bc+ca}\)
ta có: \(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\ge\dfrac{3}{2}\)(BĐT Nesbit)
\(\Rightarrow2\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)\ge\dfrac{2.3}{2}=3\)
Dấu "=" xảy ra khi \(a=b=c\)
Ta cm \(\dfrac{a^2+b^2+c^2}{ab+bc+ca}\ge1\)
Thật vậy: \(\dfrac{a^2+b^2+c^2}{ab+bc+ca}\ge1\\ \Rightarrow a^2+b^2+c^2\ge ab+bc+ca\\ \Rightarrow2a^2+2b^2+2c^2\ge2ab+2bc+2ca\\ \Rightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ac+a^2\right)\ge0\\ \Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)
Dấu "=" xảy ra khi \(a=b=c\)
\(\Rightarrow\dfrac{2a}{b+c}+\dfrac{2b}{c+a}+\dfrac{2c}{a+b}+\dfrac{a^2+b^2+c^2}{ab+bc+ca}\ge3+1=4\)
Dấu "=" xảy ra khi \(a=b=c\)
