\(P\ge\frac{\left(x+y\right)^2}{2\left(2x^2+1\right)\left(2y^2+1\right)}+\frac{1}{xy}=\frac{2}{\left(2x^2+1\right)\left(2y^2+1\right)}+\frac{2}{9xy}+\frac{7}{9xy}\)

\(P\ge\frac{8}{4x^2y^2+2x^2+2y^2+4xy+5xy+1}+\frac{7}{9xy}\)

\(P\ge\frac{8}{4\left(\frac{x+y}{2}\right)^4+2\left(x+y\right)^2+\frac{5}{4}\left(x+y\right)^2+1}+\frac{28}{9\left(x+y\right)^2}=\frac{11}{9}\)

Ta có: \(\left(2x-y\right)^2\ge0\); \(\left(y-2\right)^2\ge0\); \(\sqrt{\left(x+y+z\right)^2}=\left|x+y+z\right|\ge0\)

\(\Rightarrow\left\{{}\begin{matrix}2x-y=0\\y-2=0\\x+y+z=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=?\\y=?\\z=?\end{matrix}\right.\)

Bạn tự giải :D

Bài 4:

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\frac{x}{y+z+1}=\frac{y}{x+z+2020}=\frac{z}{x+y-2021}=\frac{x+y+z}{y+z+1+x+z+2020+x+y-2021}=\frac{x+y+z}{2\left(x+y+z\right)}=\frac12\)

=>y+z+1=2x; x+z+2020=2y; x+y-2021=z

=>y+z=2x-1; x+z=2y-2020; x+y=z+2021

\(\frac{x}{y+z+1}=x+y+z\)

=>\(x+y+z=\frac12\)

=>\(x+2x-1=\frac12\)

=>\(3x=\frac32\)

=>x=1/2

x+y+z=1/2

=>\(2y-2020+y=\frac12\)

=>3y=1/2+2020=2020,5

=>y=673,5

x+y+z=1/2

=>z+2021+z=1/2

=>2z=1/2-2021=-2020,5

=>z=-1010,25