Question

Cho 3 số thực dương a,b,c thỏa mãn: \(\frac{1}{a}+\frac{2}{b}+\frac{3}{c}=3\). CMR: \(\frac{27a^2}{c\left(c^2+9a^2\right)}+\frac{b^2}{a\left(4a^2+b^2\right)}+\frac{8c^2}{b\left(9b^2+4c^2\right)}\ge\frac{3}{2}\)

Answer 1

\(\left(\frac{1}{a};\frac{2}{b};\frac{3}{c}\right)=\left(x;y;z\right)\Rightarrow\left\{{}\begin{matrix}x+y+z=3\\a=\frac{1}{x};b=\frac{2}{y};c=\frac{3}{z}\end{matrix}\right.\)

\(VT=\frac{z^3}{z^2+x^2}+\frac{x^3}{x^2+y^2}+\frac{y^3}{y^2+z^2}\)

Ta có: \(\frac{z^3}{z^2+x^2}\ge\frac{2z-x}{2}\)

Thật vậy, BĐT tương đương \(2z^3\ge\left(2z-x\right)\left(z^2+x^2\right)\)

\(\Leftrightarrow x\left(x-z\right)^2\ge0\) (đúng)

Do đó: \(VT\ge\frac{2z-x}{2}+\frac{2x-y}{2}+\frac{2y-z}{2}=\frac{1}{2}\left(x+y+z\right)=\frac{3}{2}\)

Dấu "=" xảy ra khi \(x=y=z=1\) hay \(\left(a;b;c\right)=\left(1;2;3\right)\)