Question

Cho a, b, c dương thỏa mãn: a + b + c = 3. Chứng minh rằng:\(\dfrac{a^2+bc}{b+ca}+\dfrac{b^2+ca}{c+ab}+\dfrac{c^2+ab}{a+bc}\ge3\)

Answer 1

BĐT cần chứng minh tương đương:

\(\dfrac{a^2+bc}{3b+3ca}+\dfrac{b^2+ca}{3c+3ab}+\dfrac{c^2+ab}{3a+3bc}\ge1\)

Đặt vế trái là P

\(P=\dfrac{a^2+bc}{b\left(a+b+c\right)+3ca}+\dfrac{b^2+ca}{c\left(a+b+c\right)+3ab}+\dfrac{c^2+ab}{\left(a+b+c\right)a+3bc}\)

\(P=\dfrac{a^2+bc}{b^2+2ca+ab+bc+ca}+\dfrac{b^2+ca}{c^2+2ab+ab+bc+ca}+\dfrac{c^2+ab}{a^2+2bc+ab+bc+ca}\)

\(P\ge\dfrac{a^2+bc}{b^2+a^2+c^2+ab+bc+ca}+\dfrac{b^2+ca}{c^2+a^2+b^2+ab+bc+ca}+\dfrac{c^2+ab}{a^2+b^2+c^2+ab+bc+ca}=1\)

(đpcm)

Dấu "=" xảy ra khi \(a=b=c=1\)