Question

cho a,b,c > o thỏa mãn ab + bc + ca = 3. Cmr: \(\frac{a^3}{b^2+3}+\frac{b^3}{c^2+3}+\frac{c^3}{a^2+3}\ge\frac{3}{4}\)

Answer 1

\(3=ab+bc+ca\le\frac{\left(a+b+c\right)^2}{3}\Rightarrow a+b+c\ge3\)

Ta có: \(\frac{a^3}{b^2+3}=\frac{a^3}{b^2+ab+bc+ca}=\frac{a^3}{\left(a+b\right)\left(b+c\right)}\)

Mặt khác \(\frac{a^3}{\left(a+b\right)\left(b+c\right)}+\frac{a+b}{8}+\frac{b+c}{8}\ge\frac{3a}{4}\)

Tương tự: \(\frac{b^3}{c^3+3}+\frac{a+c}{8}+\frac{b+c}{8}\ge\frac{3b}{4}\) ; \(\frac{c^3}{a^2+8}+\frac{a+b}{8}+\frac{a+c}{8}\ge\frac{3c}{4}\)

Cộng vế với vế:

\(P+\frac{1}{2}\left(a+b+c\right)\ge\frac{3}{4}\left(a+b+c\right)\)

\(\Rightarrow P\ge\frac{1}{4}\left(a+b+c\right)\ge\frac{3}{4}\)

Dấu "=" xảy ra khi \(a=b=c=1\)