Question

Cho x,y,z>0 :xyz=1

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

Answer 1

lớp 10 rồi ....... khá là khó 

Answer 2

\(x^2+2y^2+3=x^2+y^2+y^2+1+2\ge2xy+2y+2\)

\(z^2+2x^2+3\ge2zx+2x+2\)

\(y^2+2z^2+3\ge2yz+2z+2\)

Dễ chứng minh được \(\dfrac{1}{xy+y+1}+\dfrac{1}{yz+z+1}+\dfrac{1}{zx+x+1}=1\)

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

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

Đẳng thức xảy ra khi \(x=y=z=1\)