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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).
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\Leftrightarrow xy+yz+zx=0\)
\(\Rightarrow yz=-xy-zx\Rightarrow\dfrac{yz}{x^2+2yz}=\dfrac{yz}{x^2+yz-xy-zx}=\dfrac{yz}{\left(x-y\right)\left(x-z\right)}\)
Tương tự: \(\dfrac{xz}{y^2+2xz}=\dfrac{xz}{\left(y-x\right)\left(y-z\right)}\) ; \(\dfrac{xy}{z^2+2xy}=\dfrac{xy}{\left(x-z\right)\left(y-z\right)}\)
\(\Rightarrow A=\dfrac{-yz\left(y-z\right)-zx\left(z-x\right)-xy\left(x-y\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=1\)
Bài này ez thôi, làm mãi rồi.
Theo đề bài, ta có: \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\)
=>\(\dfrac{xy+yz+xz}{xyz}=0\)
=> xy+yz+zx=0
=> \(\left\{{}\begin{matrix}xy=-yz-zx\\yz=-xy-zx\\zx=-xy-yz\end{matrix}\right.\)
Ta có: x2+2yz=x2+yz-xy-zx=(x-y)(x-z)
y2+2xz=y2+xz-xy-yz=(x-y)(z-y)
z2+2xy=z2+xy-yz-xz=(x-z)(y-z)
=> \(\dfrac{yz}{\left(x-y\right)\left(x-z\right)}+\dfrac{xz}{\left(x-y\right)\left(z-y\right)}+\dfrac{xy}{\left(x-z\right)\left(y-z\right)}=\dfrac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=\dfrac{\left(x-y\right)\left(x-z\right)\left(y-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=1\)
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:
$P=\frac{yz}{(x-z)(x-y)}+\frac{xz}{(y-z)(y-x)}+\frac{xy}{(z-x)(z-y)}$
$=\frac{-yz(y-z)-xz(z-x)-xy(x-y)}{(x-y)(y-z)(z-x)}=\frac{-[yz(y-z)+xz(z-x)+xy(x-y)]}{-[xy(x-y)+yz(y-z)+xz(z-x)]}=1$
Áp dụng Bất đẳng thức: \(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\) (Tự chứng minh)
\(\Rightarrow C=\frac{1}{x^2+2yz}+\frac{1}{y^2+2xz}+\frac{1}{z^2+2xy}\ge\frac{9}{x^2+y^2+z^2+2xy+2yz+2xz}=\frac{9}{\left(x+y+z\right)^2}\ge\frac{9}{3^2}=1\)Dấu "=" xảy ra \(\Leftrightarrow x=y=z=1\)
\(C=\frac{1}{x^2+2yz}+\frac{1}{y^2+2xz}+\frac{1}{z^2+2xy}\ge\frac{9}{x^2+y^2+z^2+2xy+2yz+2zx}=\frac{9}{\left(x+y+z\right)^2}\ge\frac{9}{3^2}=1\)
Dấu "=" xảy ra khi \(x=y=z=1\)
Xét A= \(\dfrac{x}{\sqrt{x+2yz}}\).\(\dfrac{1}{\sqrt{2}}\)=\(\dfrac{x}{\sqrt{2x+4yz}}\)=\(\sqrt{\dfrac{x.x}{2x+4yz}}\)
ta có x+y+z=\(\dfrac{1}{2}\)=> 2x+2y+2z= 1=> 2x+4yz= 4yz+1-2y-2z=(2y-1)(2z-1)
từ đó A= \(\sqrt{\dfrac{x}{2y-1}.\dfrac{x}{2z-1}}\)=\(\sqrt{\dfrac{x}{2y-2x-2y-2z}.\dfrac{x}{2z-2x-2y-2z}}\)
=\(\sqrt{\dfrac{x}{-2\left(x+y\right)}\dfrac{x}{-2\left(x+z\right)}}\)=\(\sqrt{\dfrac{1}{4}.\dfrac{x}{x+z}.\dfrac{x}{x+y}}\)=\(\dfrac{1}{2}\sqrt{\dfrac{x}{x+y}.\dfrac{x}{x+z}}\)
Áp dụng cô si \(\sqrt{ab}\)≤\(\dfrac{a+b}{2}\) =>\(\dfrac{1}{2}\sqrt{ab}\)≤\(\dfrac{a+b}{4}\)ta được
A≤\(\dfrac{1}{4}\).(\(\dfrac{x}{x+y}\)+\(\dfrac{x}{x+z}\))
cmmt thì \(\dfrac{P}{\sqrt{2}}\)≤ \(\dfrac{1}{4}\).\(\left(\dfrac{x}{x+y}+\dfrac{x}{x+z}+\dfrac{y}{y+x}+\dfrac{y}{y+z}+\dfrac{z}{z+x}+\dfrac{z}{z+y}\right)\)
\(\dfrac{P}{\sqrt{2}}\)≤\(\dfrac{3}{4}\)=>P≤\(\dfrac{3.\sqrt{2}}{4}\)=\(\dfrac{3}{2\sqrt{2}}\)
Dấu"=" xảy ra <=> x=y=z=\(\dfrac{1}{6}\)
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{0\Rightarrow\left(yz+xz+xy\right)}{xyz}=0\Rightarrow xy+xz+xy=0\)
ta có x2+2yz=x2+yz+yz=x2-yz-zx-xy=x.(x-z)-y.(x-z)=(x-y).(x-z)
tương tự ta có:x2+2xy=(x-z)*(y-z)
vậy\(A=\dfrac{yz}{\left(x-y\right).\left(x-z\right)}+\dfrac{xz}{\left(x-y\right)\left(z-y\right)}+\dfrac{xy}{\left(x-z\right)\left(y-z\right)}\)a
\(A=\dfrac{yz\left(y-z\right)}{\left(x-z\right)\left(y-z\right)\left(x-y\right)}-\dfrac{xz\left(x-z\right)}{\left(x-z\right)\left(y-z\right)\left(x-y\right)}+\dfrac{xy\left(x-y\right)}{\left(x-z\right)\left(y-z\right)\left(x-y\right)}\)
\(=\dfrac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{\left(y-z\right)\left(x-y\right)\left(x-z\right)}\)
\(=\dfrac{\left(yz-xz\right)\left(y-z\right)+\left(xy-xz\right)\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}\)
\(=\dfrac{\left(x-y\right)\left(x-z\right)\left(y-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=1\)
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\)
\(\Leftrightarrow yz+zx+xy=0\)
\(\Leftrightarrow\left[{}\begin{matrix}yz=-zx-xy\\zx=-xy-yz\\xy=-yz-zx\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{1}{x^2+2yz}=\dfrac{1}{x^2-xz-xy+yz}=\dfrac{1}{\left(x-y\right)\left(x-z\right)}\)
CMTT\(\Rightarrow\dfrac{1}{y^2+2zx}=\dfrac{1}{\left(y-z\right)\left(y-x\right)}\)
\(\dfrac{1}{z^2+2xy}=\dfrac{1}{\left(z-x\right)\left(z-y\right)}\)
\(\Rightarrow A=\dfrac{1}{\left(x-y\right)\left(x-z\right)}+\dfrac{1}{\left(y-z\right)\left(y-x\right)}+\dfrac{1}{\left(z-x\right)\left(z-y\right)}\)
\(A=\dfrac{y-z}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}+\dfrac{z-x}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}+\dfrac{x-y}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}\)
\(A=\dfrac{y-z+z-x+x-y}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=0\left(đpcm\right)\)