Question

Cho các só dương x, y, z thỏa mãn xyz = 1. Tìm giá trị lớn nhất của :

\(P=\dfrac{1}{x^2+2y^2+3}+\dfrac{1}{y^2+2z^2+3}+\dfrac{1}{z^2+2x^2+3}\)

Answer 1

Áp dụng BĐT Côsi ta có:

\(\left\{{}\begin{matrix}x^2+y^2\ge2xy\\y^2+1\ge2y\end{matrix}\right.\) \(\Rightarrow x^2+2y^2+3\ge2\left(xy+y+1\right)\)

\(\Rightarrow\dfrac{1}{x^2+2y^2+3}\le\dfrac{1}{2\left(xy+y+1\right)}\)

Tương tự cho 2 BĐT trên rồi cộng theo vế:

\(P\le\dfrac{1}{2}\left(\dfrac{1}{xy+y+1}+\dfrac{1}{yz+z+1}+\dfrac{1}{xz+x+1}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{xyz}{xy+y+xyz}+\dfrac{x}{xyz+xz+x}+\dfrac{1}{xz+x+1}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{xz}{xz+x+1}+\dfrac{x}{xz+x+1}+\dfrac{1}{xz+x+1}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{xz+x+1}{xz+x+1}=\dfrac{1}{2}\)

\("="\Leftrightarrow x=y=z=1\)