Question

a/(3a^2 + 2b^2 + c^2) + b/(3b^2 + 2c^2 + a^2) + c/(3c^2 + 2a^2 + b^2) <= 1/6(1/a + 1/b + 1/c)

Answer 1

Điều kiện a;b;c dương, đặt vế trái là P

\(P=\sum\frac{a}{a^2+c^2+2\left(a^2+b^2\right)}\le\sum\frac{a}{2ac+4ab}=\frac{1}{2}\sum\frac{1}{c+2b}=\frac{1}{2}\sum\frac{1}{b+b+c}\)

\(\Rightarrow P\le\frac{1}{18}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}+\frac{1}{c}+\frac{1}{a}\right)=\frac{1}{6}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

Dấu "=" xảy ra khi \(a=b=c\)