\(x\left(x-z\right)+y\left(y-z\right)=0\)\(\Leftrightarrow\)\(x^2+y^2=z\left(x+y\right)\)

\(\frac{x^3}{z^2+x^2}=x-\frac{z^2x}{z^2+x^2}\ge x-\frac{z^2x}{2zx}=x-\frac{z}{2}\)

\(\frac{y^3}{y^2+z^2}=y-\frac{yz^2}{y^2+z^2}\ge y-\frac{yz^2}{2yz}=y-\frac{z}{2}\)

\(\frac{x^2+y^2+4}{x+y}=\frac{z\left(x+y\right)+4}{x+y}=z-x-y+\frac{4}{x+y}+x+y\ge z-x-y+4\)

Cộng lại ra minP=4, dấu "=" xảy ra khi \(x=y=z=1\)

\(x+y+2=4xy\Rightarrow x+y+2\le\left(x+y\right)^2\)

\(\Leftrightarrow\left(x+y\right)^2-\left(x+y\right)-2\ge0\)

\(\Leftrightarrow\left(x+y-2\right)\left(x+y+1\right)\ge0\)

\(\Leftrightarrow x+y-2\ge0\) (do x+y+1>0 với mọi x,y>0)

\(\Leftrightarrow x+y\ge2\)

Có \(x+y+\dfrac{1}{x+y}=\left(x+y\right)+\dfrac{4}{x+y}-\dfrac{3}{x+y}\)\(\ge2\sqrt{\left(x+y\right).\dfrac{4}{x+y}}-\dfrac{3}{2}=\dfrac{5}{2}\)

Dấu = xảy ra <=> x=y=1

Vậy GTNN của biểu thức là \(\dfrac{5}{2}\)

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