Question

Cho a, b, c > 0 và ab + bc + ca = 1.

CMR : \(\dfrac{a^3}{b+c}+\dfrac{b^3}{a+c}+\dfrac{c^3}{a+b}>\dfrac{1}{2}\)

Answer 1

Ta có :

\(VT=\dfrac{a^3}{b+c}+\dfrac{b^3}{a+c}+\dfrac{c^3}{a+b}=\dfrac{a^4}{ab+ac}+\dfrac{b^4}{bc+ab}+\dfrac{c^4}{ac+bc}\)

Theo BĐT Cauchy ta có :

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

Theo BĐT Cô - Si ta lại có : \(a^2+b^2+c^2\ge ab+bc+ac\)

\(\Rightarrow VT\ge\dfrac{\left(ab+bc+ca\right)^2}{2\left(ab+bc+ca\right)}=\dfrac{ab+bc+ca}{2}=\dfrac{1}{2}\)