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Đẳng thức đã cho tương đương với:
\(\dfrac{x^2z+y^2z-z^3+y^2x+z^2x-x^3+z^2y+x^2y-y^3}{2yxz}=1\)
\(\Leftrightarrow x^3+y^3+z^3+2xyz-x^2y-y^2z-z^2x-xy^2-yz^2-zx^2=0\)
\(\Leftrightarrow\left(x+y-z\right)\left(y+z-x\right)\left(z+x-y\right)=0\Leftrightarrow z+x=y\) (Do x + y khác z và y + z khác x).
Từ đó P = 2y (Biểu thức của P phụ thuộc vào biến y).
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 : \(x=y=z=\dfrac{1}{3}\)
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\Leftrightarrow xy+yz+zx=0\)
\(\Rightarrow yz=-xy-zx\Rightarrow\dfrac{yz}{x^2+2yz}=\dfrac{yz}{x^2+yz-xy-zx}=\dfrac{yz}{\left(x-y\right)\left(x-z\right)}\)
Tương tự: \(\dfrac{xz}{y^2+2xz}=\dfrac{xz}{\left(y-x\right)\left(y-z\right)}\) ; \(\dfrac{xy}{z^2+2xy}=\dfrac{xy}{\left(x-z\right)\left(y-z\right)}\)
\(\Rightarrow A=\dfrac{-yz\left(y-z\right)-zx\left(z-x\right)-xy\left(x-y\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=1\)
Lời giải:
Từ $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$
$\Rightarrow xy+yz+xz=0$
Khi đó:
$x^2+2yz=x^2+yz-xz-xy=(x^2-xy)-(xz-yz)=x(x-y)-z(x-y)=(x-z)(x-y)$
Tương tự với $y^2+2zx, z^2+2xy$ thì:
$P=\frac{yz}{(x-z)(x-y)}+\frac{xz}{(y-z)(y-x)}+\frac{xy}{(z-x)(z-y)}$
$=\frac{-yz(y-z)-xz(z-x)-xy(x-y)}{(x-y)(y-z)(z-x)}=\frac{-[yz(y-z)+xz(z-x)+xy(x-y)]}{-[xy(x-y)+yz(y-z)+xz(z-x)]}=1$