Đặt \(A = \frac{y^2 + z^2 - x^2}{2yz}, \quad B = \frac{z^2 + x^2 - y^2}{2xz}, \quad C = \frac{x^2 + y^2 - z^2}{2xy}\)

\(1+A=\frac{y^2+z^2-x^2}{2yz}+1=\frac{y^2+2yz+z^2-x^2}{2yz}=\frac{\left(y+z\right)^2-x^2}{2yz}=\frac{\left(y+z-x\right)\left(y+z+x\right)}{2yz}\)

\(1 + B = \frac{(z + x - y)(x + y + z)}{2xz}\)

\(1 + C = \frac{(x + y - z)(x + y + z)}{2xy}\)

\(1-A=1-\frac{y^2+z^2-x^2}{2yz}=\frac{x^2-\left(y^2-2yz+z^2\right)}{2yz}\)

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

\(1 - B = \frac{(y - z + x)(y + z - x)}{2xz}\)

\(1 - C = \frac{(z - x + y)(z + x - y)}{2xy}\)

(1-A)(1-B)(1-C)

\(=\frac{(x + y - z)(y + z - x)(z + x - y) \cdot(x - y + z)(y - z + x)(z - x + y)}{8x^2 y^2 z^2}\)

\(= \frac{(x + y - z)^2 (y + z - x)^2 (z + x - y)^2}{8x^2 y^2 z^2}\)

(1+A)(1+B)(1+C)

\(=\frac{\left(x+y+z\right)\cdot\left(x+y+z\right)\left(x+y+z\right)(y+z-x)\cdot\left.(z+x-y\right)(x+y-z)}{8x^2y^2z^2}\)

\(=\frac{(x + y + z)^3 (y + z - x)(z + x - y)(x + y - z)}{8x^2 y^2 z^2}\)

Theo đề, ta có: A+B+C=1

=>\(x(y^2 + z^2 - x^2) + y(z^2 + x^2 - y^2) + z(x^2 + y^2 - z^2) = 2xyz\)

=>\(xy^2 + xz^2 - x^3 + yz^2 + yx^2 - y^3 + zx^2 + zy^2 - z^3 = 2xyz\)

=>\(-(x^3 + y^3 + z^3) + (xy^2 + x^2y) + (yz^2 + y^2z) + (zx^2 + z^2x) - 2xyz = 0\)

=>\((x + y - z)(y + z - x)(z + x - y) = 0\)

TH1: x+y-z=0

=>z=x+y

\(A = \frac{y^2 + (x+y)^2 - x^2}{2y(x+y)} = \frac{y^2 + x^2 + 2xy + y^2 - x^2}{2y(x+y)} = \frac{2y^2 + 2xy}{2y(x+y)} = \frac{2y(x+y)}{2y(x+y)} = 1\)

\(B = \frac{(x+y)^2 + x^2 - y^2}{2x(x+y)} = \frac{x^2 + 2xy + y^2 + x^2 - y^2}{2x(x+y)} = \frac{2x^2 + 2xy}{2x(x+y)} = \frac{2x(x+y)}{2x(x+y)} = 1\)

\(C = \frac{x^2 + y^2 - (x+y)^2}{2xy} = \frac{x^2 + y^2 - (x^2 + 2xy + y^2)}{2xy} = \frac{-2xy}{2xy} = -1\)

=>A=B=1; C=-1(1)

TH2: y+z-x=0

=>x=y+z

\(A = \frac{y^2 + z^2 - (y+z)^2}{2yz} = \frac{-2yz}{2yz} = -1\)

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

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

Do đó: B=C=1; A=-1(2)

TH3: z+x-y=0

=>y=x+z

\(A = \frac{y^2 + z^2 - x^2}{2yz} = \frac{(x+z)^2 + z^2 - x^2}{2(x+z)z}\)

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

\(B = \frac{z^2 + x^2 - y^2}{2xz} = \frac{z^2 + x^2 - (x+z)^2}{2xz}\)

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

\(C = \frac{x^2 + y^2 - z^2}{2xy} = \frac{x^2 + (x+z)^2 - z^2}{2x(x+z)}\)

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

Do đó: A=C=1; B=-1(3)

Từ (1),(2),(3) suy ra trong 3 phân thức A,B,C; sẽ có hai phân thức bằng 1 và phân thức còn lại bằng -1

Lời giải:

\(\frac{x^2+y^2-z^2}{2xy}+\frac{y^2+z^2-x^2}{2yz}+\frac{x^2+z^2-y^2}{2xz}=1\)

\(\Leftrightarrow \frac{x^2+y^2-z^2}{2xy}+1+\frac{y^2+z^2-x^2}{2yz}-1+\frac{x^2+z^2-y^2}{2xz}-1=0\)

\(\Leftrightarrow \frac{(x+y-z)(x+y+z)}{2xy}+\frac{(y-z-x)(y-z+x)}{2yz}+\frac{(x-z-y)(x-z+y)}{2xz}=0\)

\(\Leftrightarrow (x+y-z)\left[\frac{x+y+z}{2xy}+\frac{y-z-x}{2yz}+\frac{x-z-y}{2xz}\right]=0\)

\(\Leftrightarrow (x+y-z)(xz+yz+z^2+xy-zx-x^2+xy-zy-y^2)=0\)

\(\Leftrightarrow (x+y-z)[z^2-(x-y)^2]=0\Leftrightarrow (x+y-z)(z-x+y)(x+z-y)=0\)

Nếu $x+y-z=0$ thì:

\(\frac{x^2+y^2-z^2}{2xy}=\frac{(x+y)^2-z^2-2xy}{2xy}=-1\); \(\frac{y^2+z^2-x^2}{2yz}=\frac{z(y-x)+z^2}{2yz}=\frac{y-x+z}{2y}=\frac{y-x+y+x}{2y}=1\)

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

Ta có đpcm.

Các TH còn lại tương tự.

Vậy........

Biến thì khác nhau nhưng quan trọng là cách làm :)) 

Vào TKHĐ của tớ để xem hình ảnh nhé, dài ngại chả muốn viết :V

Ta có :

\(x+y+z=1\)

\(\Rightarrow\left(x+y+z\right)^2=1\)

Áp dụng BĐT Cauchy-schwar dưới dạng engel ta có :

\(\dfrac{1}{x^2+2yz}+\dfrac{1}{y^2+2zx}+\dfrac{1}{z^2+2xy}\ge\dfrac{\left(1+1+1\right)^2}{x^2+y^2+z^2+2xy+2yz+2zx}=\dfrac{9}{1}=9\)

\(\text{Ta có : }x+y+z=1\\ \Rightarrow\left(x+y+z\right)^2=1\\ \Rightarrow x^2+y^2+z^2+2xy+2xz+2yz=1\\ \Rightarrow\dfrac{1}{x^2+2yz}+\dfrac{1}{y^2+2xz}+\dfrac{1}{z^2+2xy}\\ =\dfrac{x^2+y^2+z^2+2xy+2xz+2yz}{x^2+2yz}+\dfrac{x^2+y^2+z^2+2xy+2xz+2yz}{y^2+2xz}+\dfrac{x^2+y^2+z^2+2xy+2xz+2yz}{z^2+2xy}\\ =\dfrac{x^2+2yz}{x^2+2yz}+\dfrac{y^2+2xz}{x^2+2yz}+\dfrac{z^2+2xy}{x^2+2yz}+\dfrac{x^2+2yz}{y^2+2xz}+\dfrac{y^2+2xz}{y^2+2xz}+\dfrac{z^2+2xy}{y^2+2xz}+\dfrac{x^2+2yz}{z^2+2xy}+\dfrac{y^2+2xz}{z^2+2xy}+\dfrac{z^2+2xy}{z^2+2xy}\\ =1+\left(\dfrac{y^2+2xz}{x^2+2yz}+\dfrac{x^2+2yz}{y^2+2xz}\right)+\left(\dfrac{z^2+2xy}{x^2+2yz}+\dfrac{x^2+2yz}{z^2+2xy}\right)+1+\left(\dfrac{y^2+2xz}{z^2+2xy}+\dfrac{z^2+2xy}{y^2+2xz}\right)+1\)Áp dụng \(BDT:\dfrac{a}{b}+\dfrac{b}{a}\ge2\)

\(\Rightarrow1+\left(\dfrac{y^2+2xz}{x^2+2yz}+\dfrac{x^2+2yz}{y^2+2xz}\right)+\left(\dfrac{z^2+2xy}{x^2+2yz}+\dfrac{x^2+2yz}{z^2+2xy}\right)+1+\left(\dfrac{y^2+2xz}{z^2+2xy}+\dfrac{z^2+2xy}{y^2+2xz}\right)+1\\ \ge1+2+2+1+2+1\ge9\left(đpcm\right)\)

Dấu \("="\) xảy ra khi: \(\left\{{}\begin{matrix}y^2+2xz=x^2+2yz\\z^2+2xy=x^2+2yz\\y^2+2xz=z^2+2xy\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y^2-2yz=x^2-2xz\\z^2-2yz=x^2-2xy\\y^2-2xy=z^2-2xz\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y^2-2yx+z^2=x^2-2xz+z^2\\z^2-2yz+y^2=x^2-2xy+y^2\\y^2-2xy+x^2=z^2-2xz+x^2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\left(y-z\right)^2=\left(x-z\right)^2\\\left(z-y\right)^2=\left(x-y\right)^2\\\left(y-x\right)^2=\left(z-x\right)^2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y-z=x-z\\z-y=x-y\\y-x=z-x\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=y\\z=x\\y=z\end{matrix}\right.\Leftrightarrow x=y=z\\\text{Mà } x+y+z=1\\ \Leftrightarrow3x=1\\ \Leftrightarrow x=\dfrac{1}{3}\\ \Leftrightarrow x=y=z=\dfrac{1}{3}\)

Vậy \(\dfrac{1}{x^2+2yz}+\dfrac{1}{y^2+2xz}+\dfrac{1}{z^2+2xy}\ge9\) với \(x;y;z>0\) và \(x+y+z=1\)

đẳng thức xảy ra khi : \(x=y=z=\dfrac{1}{3}\)

Lời giải:

Từ \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Rightarrow \frac{xy+yz+xz}{xyz}=0\Rightarrow xy+yz+xz=0\)

Suy ra \(yz=-xy-xz\)

\(\Rightarrow x^2+2yz=x^2+yz-xy-xz=x(x-y)-z(x-y)\)

\(\Leftrightarrow x^2+2yz=(x-z)(x-y)\)

\(\Rightarrow \frac{yz}{x^2+2yz}=\frac{yz}{(x-z)(x-y)}\)

Hoàn toàn tương tự với các phân thức còn lại và cộng theo vế:

\(A=\frac{yz}{(x-y)(x-z)}+\frac{xz}{(y-x)(y-z)}+\frac{xy}{(z-x)(z-y)}\)

\(A=\frac{-yz(y-z)}{(x-y)(y-z)(z-x)}+\frac{-xz(z-x)}{(x-y)(y-z)(z-x)}+\frac{-xy(x-y)}{x-y)(y-z)(z-x)}\)

\(A=\frac{xy^2+yz^2+zx^2-(x^2y+y^2z+z^2x)}{(x-y)(y-z)(z-x)}\)

\(A=\frac{xy^2+yz^2+zx^2-(x^2y+y^2z+z^2x)}{xy^2+yz^2+zx^2-(x^2y+y^2z+z^2x)}=1\)

Ta có: \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\)\(\Rightarrow xy+yz+xz=0\)

\(\Rightarrow\left\{{}\begin{matrix}xy=-yz-xz\\yz=-xy-xz\\xz=-xy-xz\end{matrix}\right.\)

\(\Rightarrow\dfrac{yz}{x^2+2yz}=\dfrac{yz}{x^2+yz-xy-xz}=\dfrac{yz}{\left(x-y\right)\left(x-z\right)}\)

Tương tự:

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{xz}{y^2+2xz}=\dfrac{xz}{\left(x-y\right)\left(x-z\right)}\\\dfrac{xy}{z^2+2xy}=\dfrac{xy}{\left(x-y\right)\left(x-z\right)}\\\dfrac{yz}{x^2+2yz}=\dfrac{yz}{\left(x-y\right)\left(x-z\right)}\end{matrix}\right.\)

\(\Rightarrow A=\dfrac{xz}{\left(x-y\right)\left(x-z\right)}+\dfrac{xy}{\left(x-y\right)\left(x-z\right)}+\dfrac{yz}{\left(x-y\right)\left(x-z\right)}=\dfrac{xz+xy+yz}{\left(x-y\right)\left(x-z\right)}=\dfrac{0}{\left(x-y\right)\left(x-z\right)}=0\)

Vậy \(A=0.\)