Question

cho a;b;c>0 thỏa mãn a+b+c=1.Tìm Max của bt:

\(A=\frac{a}{9a^3+3b^2+c}+\frac{b}{9b^3+3c^2+a}+\frac{c}{9c^3+3a^2+b}\)

Answer 1

Áp dụng BĐT AM-GM ta có:

\(9a^3+\frac{1}{3}+\frac{1}{3}\ge3\sqrt[3]{9a^3\cdot\frac{1}{3}\cdot\frac{1}{3}}=3a\)

\(3b^2+\frac{1}{3}\ge2\sqrt{3b^2\cdot\frac{1}{3}}=2b\)

Do đó: \(A\le\text{∑}\frac{a}{3a+2b+c-1}=\frac{a}{2a+b}\left(a+b+c=1\right)\)

\(2A\le\text{∑}\frac{2a}{2a+b}=3-\text{∑}\frac{b}{2a+b}=3-\text{∑}\frac{b^2}{2ab+b^2}\)

Áp dụng BĐT Cauchy-Schwarz ta có:

\(2A\le3-\frac{\left(a+b+c\right)^2}{a^2+b^2+c^2+2ab+2bc+2ca}\)

\(=3-\frac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2}=2\Leftrightarrow A\le1\)

Dấu "=" khi \(a=b=c=\frac{1}{3}\)