x^2+1>=2x suy ra 1/x^2+1=y<=1/2x+y=1/x+x+y=1/9(9/x+x+y)<=1/x+1/x+1/y.

A(BT)<=1/9(3/x+3/y+3/z)=1/3(1/x+1/y+1/z)

Mà từ x+y+z=xy+yz+zx suy ra x+y+z=xy+yz+zx>=3

dễ dàng cm bằng phương pháp đánh giá suy ra 1/x+1/y+1/z<3

suy ra A<1/3.3=1(đpcm)

\(x^2+xy+y^2=\left(x+y\right)^2-xy\ge\left(x+y\right)^2-\frac{1}{4}\left(x+y\right)^2=\frac{3}{4}\left(x+y\right)^2\)

\(\Rightarrow\sqrt{x^2+xy+y^2}\ge\frac{\sqrt{3}}{2}\left(x+y\right)\)

Vậy:

\(P\ge\frac{\sqrt{3}}{2}\left[\frac{\left(x+y\right)^2}{1+4xy}+\frac{\left(y+z\right)^2}{1+4yz}+\frac{\left(z+x\right)^2}{1+4zx}\right]\)

\(P\ge\frac{\sqrt{3}}{2}\left[\frac{\left(2x+2y+2z\right)^2}{3+4\left(xy+yz+zx\right)}\right]\ge\frac{\sqrt{3}}{2}.\frac{9}{3+\frac{4}{3}\left(x+y+z\right)^2}=\frac{3\sqrt{3}}{4}\)

Dấu "=" xảy ra khi \(x=y=z=\frac{1}{2}\)

Câu hỏi của Vũ Thảo Vy - Toán lớp 8 - Học toán với OnlineMath tham khảo

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Leftrightarrow xy+yz+zx=0\)

\(A=\frac{yz}{x^2}+\frac{xz}{y^2}+\frac{xy}{z^2}=\frac{y^3z^3+x^3z^3+x^3y^3}{x^2y^2z^2}=\frac{\left(xy+yz+xz\right)\left(...\right)}{x^2y^2z^2}=0\)

Đặt bài toán phụ : Chứng minh nếu \(a+b+c=0\Rightarrow a^3+b^3+c^3=3abc\)

Thật vậy :

 \(a^3+b^3+c^3=\left(a+b+c\right)^3-3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)

\(a+b+c=0\Rightarrow\left(a+b+c\right)^3=0\)

\(a+b=-c\)

\(b+c=-a\)

\(c+a=-b\)

\(\Rightarrow\left(a+b+c\right)^3-3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)

\(=-3\left(-c\right)\left(-b\right)\left(-a\right)\)

\(=3abc\)

Trở lại bài toán chính :

Ta có:

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)

\(\Rightarrow\frac{yz}{xyz}+\frac{xz}{xyz}+\frac{xy}{xyz}=0\)

\(\Rightarrow\frac{yz+xz+xy}{xyz}=0\)

\(\Rightarrow xy+xz+yz=0\)

\(\Rightarrow\left(xy\right)^3+\left(xz\right)^3+\left(yz^3\right)=3\left(xy\right)\left(xz\right)\left(yz\right)=3x^2y^2z^2\)

Lại có:

\(P=\frac{xy.y^2x^2}{x^2y^2z^2}+\frac{xz.z^2.x^2}{x^2y^2z^2}+\frac{z^2.y^2.yz}{x^2y^2z^2}\)

\(=\frac{\left(xy\right)^3}{x^2y^2z^2}+\frac{\left(xz\right)^3}{x^2y^2z^2}+\frac{\left(yz\right)^3}{x^2y^2z^2}\)

\(=\frac{\left(xy\right)^3+\left(xz\right)^3+\left(yz^3\right)}{x^2y^2z^2}\)

Thay \(\left(xy\right)^3+\left(xz\right)^3+\left(yz^3\right)=3x^2y^2z^2;\)ta có:

\(P=\frac{3x^2y^2z^2}{x^2y^2z^2}\)

\(=3\)

Vậy \(P=3.\)