Áp dụng Bunhia.
\(\left(x+y+z\right)^2\le\left(1^2+1^2+1^2\right)\left(x^2+y^2+z^2\right)=3.3=9\)
=> \(0< x+y+z\le3\)
Có: \(P=\frac{x^2+1}{x}+\frac{y^2+1}{y}+\frac{z^2+1}{z}-\frac{1}{x+y+z}\)
\(=\frac{x^2-2x+1}{x}+\frac{y^2-2y+1}{y}+\frac{z^2-2z+1}{z}-\frac{1}{x+y+z}+6\)
\(=\frac{\left(x-1\right)^2}{x}+\frac{\left(y-1\right)^2}{y}+\frac{\left(z-1\right)^2}{z}-\frac{1}{x+y+z}+6\)
\(\ge\frac{\left(x+y+z-3\right)^2}{x+y+z}-\frac{1}{x+y+z}+6=\frac{\left(x+y+z-3\right)^2-1}{x+y+z}+6\)
\(\ge\frac{0-1}{3}+6=\frac{17}{3}\)
"=" xảy ra <=> \(x+y+z=3;x=y=z\Leftrightarrow x=y=z=1\)
Vậy min P = 17/3 <=> x = y = z =1.
\(P=\frac{x^2+1}{x}+\frac{y^2+1}{y}+\frac{z^2+1}{z}-\frac{1}{x+y+z}\)
\(=x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-\frac{1}{x+y+z}\)
\(\ge x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-\frac{1}{9}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
\(=x+y+z+\frac{8x}{9}+\frac{8y}{9}+\frac{8z}{9}\)
Có BĐT phụ \(a+\frac{8}{9a}\ge\frac{a^2+33}{18}\)
\(\Leftrightarrow\frac{9a^2+8}{9a}\ge\frac{a^2+33}{18}\)
\(\Leftrightarrow162a^2+144-9a^3-297a\ge0\)
\(\Leftrightarrow-a^3+18a^2-33a+16\ge0\)
\(\Leftrightarrow\left(a-1\right)^2\left(16-a\right)\ge0\left(OK\right)\)
\(\Rightarrow P\ge\frac{x^2+y^2+z^2+99}{18}=\frac{17}{3}\)
Dấu "=" xảy ra tại x=y=z=1
\(\frac{\left(x+y+z-3\right)^2-1}{x+y+z}+6\ge\frac{0-1}{x+y+z}+6\ge\frac{0-1}{3}+6\)
Chú ý: \(\left(x+y+z-3\right)^2\ge0\) với mọi x, y, z.
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\(P=x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-\frac{1}{x+y+z}\)
\(=\Sigma_{cyc}\frac{\left(16-x\right)\left(x-1\right)^2}{18x}+\Sigma_{cyc}\frac{\left(x-y\right)^2}{9xy\left(x+y+z\right)}+\frac{a^2+b^2+c^2+33.3}{18}\)
\(=\Sigma_{cyc}\frac{\left(16-x\right)\left(x-1\right)^2}{18x}+\Sigma_{cyc}\frac{\left(x-y\right)^2}{9xy\left(x+y+z\right)}+\frac{17}{3}\)
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