Question

[Nguyễn Linh Chi]

Dũng Senpai 

Cho x, y, z > 0 và \(x^2+y^2+z^2=3\)

Tìm min. \(P=\frac{x^2+1}{x}+\frac{y^2+1}{y}+\frac{z^2+1}{z}-\frac{1}{x+y+z}\)

Answer 1

[Nguyễn Linh Chi]

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.

Answer 2

[zZz Cool Kid_new zZz]

\(P=\frac{x^2+1}{x}+\frac{y^2+1}{y}+\frac{z^2+1}{z}-\frac{1}{x+y+z}\)

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

\(\ge x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-\frac{1}{9}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)

\(=x+y+z+\frac{8x}{9}+\frac{8y}{9}+\frac{8z}{9}\)

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