Question

Cho a > b > c > 0 và a² + b² + c² = 1
Chứng minh rằng \(\dfrac{a^3}{b+c}+\dfrac{b^3}{a+c}+\dfrac{c^3}{a+b}\ge\dfrac{1}{2}\)

Answer 1

Áp dụng liên tiếp bđt Cauchy-Schwarz và hệ quả AM-GM:

\(NL=\dfrac{a^3}{b+c}+\dfrac{b^3}{a+c}+\dfrac{c^3}{a+b}=\dfrac{a^4}{ab+ac}+\dfrac{b^4}{ab+bc}+\dfrac{c^4}{ac+bc}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{2\left(ab+bc+ac\right)}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{2\left(a^2+b^2+c^2\right)}=\dfrac{a^2+b^2+c^2}{2}=\dfrac{1}{2}\)

P/s: Bài này điều kiện thừa,k nên cho a>b>c

Answer 2

\(\dfrac{a^3}{b+c}+\dfrac{b^3}{a+c}+\dfrac{c^3}{a+b}\)

\(=\dfrac{a^4}{ab+ac}+\dfrac{b^4}{ab+bc}+\dfrac{c^4}{ac+bc}\)

\(\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{2\left(ab+bc+ac\right)}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{2\left(a^2+b^2+c^2\right)}\)

\(=\dfrac{a^2+b^2+c^2}{2}=\dfrac{1}{2}\)

Dấu "=" xảy ra khi: \(a=b=c=\dfrac{1}{\sqrt{3}}\)