Question

Cho ba số x;y;z thỏa mãn \(xyz=2\). Chứng minh :
\(\dfrac{x}{2x^2+y^2+5}+\dfrac{2y}{6y^2+z^2+6}+\dfrac{4z}{3z^2+4x^2+16}\le\dfrac{1}{2}\)

Answer 1

Đặt vế trái là P:

\(P=\dfrac{x}{\left(x^2+y^2\right)+\left(x^2+1\right)+4}+\dfrac{2y}{\left(4y^2+z^2\right)+2\left(y^2+1\right)+4}+\dfrac{4z}{\left(z^2+4x^2\right)+2\left(z^2+4\right)+8}\)

\(P\le\dfrac{x}{2xy+2x+4}+\dfrac{2y}{4yz+4y+4}+\dfrac{4z}{4zx+8z+8}=\dfrac{x}{2xy+2x+4}+\dfrac{y}{2yz+2y+2}+\dfrac{z}{zx+2z+2}\)

\(=\dfrac{x}{2xy+2x+4}+\dfrac{xy}{2xyz+2xy+2x}+\dfrac{xyz}{zx.xy+2xyz+2xy}\)

\(=\dfrac{x}{2xy+2x+4}+\dfrac{xy}{4+2xy+2x}+\dfrac{2}{2x+4+2xy}=\dfrac{1}{2}\) (đpcm)

Dấu "=" xảy ra khi \(\left(x;y;z\right)=\left(1;1;2\right)\)