Question

Cho x,y,z>0 thỏa mãn xyz=1.Chứng minh

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

Answer 1

Áp dụng BĐT \(AM-GM\) ta có :

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

Tương tự : \(\dfrac{1}{y^2+2z^2+3}\le\dfrac{1}{2\left(yz+z+1\right)}\)

\(\dfrac{1}{z^2+2x^2+3}\le\dfrac{1}{2\left(zx+x+1\right)}\)

Cộng từng vế BĐT ta được :

\(\dfrac{1}{x^2+2y^2+3}+\dfrac{1}{y^2+2z^2+3}+\dfrac{1}{z^2+2x^2+3}\le\dfrac{1}{2}\left(\dfrac{1}{xy+y+1}+\dfrac{1}{yz+z+1}+\dfrac{1}{zx+x+1}\right)=\dfrac{1}{2}\left(\dfrac{xyz}{xy+y+xyz}+\dfrac{x}{xyz+zx+x}+\dfrac{1}{zx+x+1}\right)=\dfrac{1}{2}\left(\dfrac{xz+x+1}{xy+x+1}\right)=\dfrac{1}{2}.1=\dfrac{1}{2}\)