Question

Cho a, b, c > 0 thỏa mãn\(a^2+b^2+c^2=1\)CMR \(\dfrac{a^3}{b^2+c}+\dfrac{b^3}{c^2+a}+\dfrac{c^3}{a^2+b}\ge\dfrac{\sqrt{3}}{1+\sqrt{3}}\)

Answer 1

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

\(\ge\dfrac{1}{\sqrt{\left(a^2+b^2+c^2\right)\left(a^2c^2+b^2c^2+a^2b^2\right)}+a^2+b^2+c^2}\ge\dfrac{1}{\sqrt{\dfrac{\left(a^2+b^2+c^2\right)^2}{3}}+1}=\dfrac{1}{\dfrac{1}{\sqrt{3}}+1}=\dfrac{\sqrt{3}}{1+\sqrt{3}}\)

\(dau"="\Leftrightarrow a=b=c=\dfrac{1}{\sqrt{3}}\)