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\(5x^2+2xy+2y^2-\left(4x^2+4xy+y^2\right)=\left(x-y\right)^2\ge0\\ \Leftrightarrow5x^2+2xy+2y^2\ge4x^2+4xy+y^2=\left(2x+y\right)^2\)
\(\Leftrightarrow P\le\dfrac{1}{2x+y}+\dfrac{1}{2y+z}+\dfrac{1}{2z+x}=\dfrac{1}{9}\left(\dfrac{9}{x+x+y}+\dfrac{9}{y+y+z}+\dfrac{9}{z+z+x}\right)\\ \Leftrightarrow P\le\dfrac{1}{9}\left(\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{y}+\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{z}+\dfrac{1}{z}+\dfrac{1}{x}\right)\\ \Leftrightarrow P\le\dfrac{1}{9}\left(\dfrac{3}{x}+\dfrac{3}{y}+\dfrac{3}{z}\right)=\dfrac{1}{3}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=1\)
Dấu \("="\Leftrightarrow x=y=z=1\)
\(P=\sum\dfrac{1}{x+y+1}\ge\dfrac{9}{2\left(x+y+z\right)+3}=\dfrac{9}{2.1+3}=\dfrac{9}{5}\)
Dấu \("="\Leftrightarrow x=y=z=\dfrac{1}{3}\)
Ta có: \(\dfrac{x}{x+1}=1-\dfrac{1}{x+1}\)
\(\dfrac{y}{y+1}=1-\dfrac{1}{y+1}\)
\(\dfrac{z}{z+1}=1-\dfrac{1}{z+1}\)
Cộng vế theo vế:
\(P=3-\left(\dfrac{1}{x+1}+\dfrac{1}{y+1}+\dfrac{1}{z+1}\right)\)
Áp dụng BĐT Cauchy- Schwarz dạng Engel:
\(\dfrac{1}{x+1}+\dfrac{1}{y+1}+\dfrac{1}{z+1}\ge\dfrac{\left(1+1+1\right)^2}{x+y+z+3}=\dfrac{9}{4}\)
\(\Rightarrow P\le3-\dfrac{9}{4}=\dfrac{3}{4}\)
\("="\Leftrightarrow x=y=z=\dfrac{1}{3}\)
