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Áp dụng BĐT Cauchy-Schwarz ta có:
\(\dfrac{x}{2x+y+z}=\dfrac{x}{\left(x+y\right)+\left(x+z\right)}\le\dfrac{1}{4}\left(\dfrac{x}{x+y}+\dfrac{x}{x+z}\right)\)
Tương tự cho 2 BĐT còn lại ta cũng có:
\(\dfrac{y}{2y+x+z}\le\dfrac{1}{4}\left(\dfrac{y}{x+y}+\dfrac{y}{y+z}\right);\dfrac{z}{2z+y+x}\le\dfrac{1}{4}\left(\dfrac{z}{y+z}+\dfrac{z}{x+z}\right)\)
Cộng theo vế 3 BĐT trên ta có:
\(VT\le\dfrac{1}{4}\left(\dfrac{x}{x+y}+\dfrac{x}{x+z}\right)+\dfrac{1}{4}\left(\dfrac{y}{x+y}+\dfrac{y}{y+z}\right)+\dfrac{1}{4}\left(\dfrac{z}{y+z}+\dfrac{z}{x+z}\right)\)
\(=\dfrac{1}{4}\left(\dfrac{x}{x+y}+\dfrac{y}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{y+z}+\dfrac{x}{x+z}+\dfrac{z}{x+z}\right)\)
\(=\dfrac{1}{4}\left(\dfrac{x+y}{x+y}+\dfrac{y+z}{y+z}+\dfrac{x+z}{x+z}\right)=\dfrac{1}{4}\left(1+1+1\right)=\dfrac{3}{4}\)
Ta có:
\(\dfrac{x}{2x+y+z}=\dfrac{x}{\left(x+y\right)+\left(y+z\right)}\le\dfrac{x}{2\sqrt{\left(x+y\right)\left(y+z\right)}}\)
Tương tự với các phân số khác
\(\Rightarrow VT\le\dfrac{1}{2}\left(\dfrac{x}{\sqrt{\left(x+y\right)\left(z+x\right)}}+\dfrac{y}{\sqrt{\left(y+z\right)\left(x+y\right)}}+\dfrac{z}{\sqrt{\left(z+x\right)\left(x+y\right)}}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{\sqrt{x}\cdot\sqrt{x}}{\sqrt{x+y}\cdot\sqrt{z+x}}+\dfrac{\sqrt{y}\cdot\sqrt{y}}{\sqrt{y+z}\cdot\sqrt{x+y}}+\dfrac{\sqrt{z}\cdot\sqrt{z}}{\sqrt{z+x}\cdot\sqrt{y+z}}\right)\)
\(\le\dfrac{1}{2}\left(\dfrac{\dfrac{x}{x+y}+\dfrac{x}{z+x}}{2}+\dfrac{\dfrac{y}{y+z}+\dfrac{y}{x+y}}{2}+\dfrac{\dfrac{z}{z+x}+\dfrac{z}{y+z}}{2}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{\left(\dfrac{x}{x+y}+\dfrac{y}{x+y}\right)+\left(\dfrac{y}{y+z}+\dfrac{z}{y+z}\right)+\left(\dfrac{z}{z+x}+\dfrac{x}{z+x}\right)}{2}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{3}{2}=\dfrac{3}{4}\)
Dấu "=" xảy ra khi x = y = z
\(\left\{{}\begin{matrix}\dfrac{x}{x+y}>\dfrac{x}{x+y+z}\\\dfrac{y}{y+z}>\dfrac{y}{x+y+z}\\\dfrac{z}{x+z}>\dfrac{z}{x+y+z}\end{matrix}\right.\)
\(\Rightarrow\dfrac{x}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{z+x}>\dfrac{x}{x+y+z}+\dfrac{y}{x+y+z}+\dfrac{z}{x+y+z}\)
\(\Rightarrow\dfrac{x}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{z+x}>1\)
\(\left\{{}\begin{matrix}\dfrac{x}{x+y}< \dfrac{x+z}{x+y+z}\\\dfrac{y}{y+z}< \dfrac{y+x}{x+y+z}\\\dfrac{z}{x+z}< \dfrac{z+y}{x+y+z}\end{matrix}\right.\)
\(\Rightarrow\dfrac{x}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{z+x}< \dfrac{x+z}{x+y+z}+\dfrac{y+x}{x+y+z}+\dfrac{z+y}{x+y+z}\)
\(\Rightarrow\dfrac{x}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{z+x}< \dfrac{x+z+y+x+z+y}{x+y+z}\)
\(\Rightarrow\dfrac{x}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{z+x}< \dfrac{2\left(x+y+z\right)}{x+y+z}=2\)
\(\Rightarrow1< \dfrac{x}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{z+x}< 2\)
TH1 : \(x+y+z=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=-z\\y+z=-x\\x+z=-y\end{matrix}\right.\)
\(\Leftrightarrow M=\dfrac{\left(-z\right)\left(-x\right)\left(-y\right)}{8xyz}=\dfrac{-\left(xyz\right)}{8xyz}=\dfrac{-1}{8}\)
Th2 : \(x+y+z\ne0\)
\(\dfrac{2x+2y-z}{z}=\dfrac{2x-2z+y}{y}=\dfrac{2y+2z-x}{x}\)
\(\Leftrightarrow\left(\dfrac{2x+2y-z}{z}+3\right)=\left(\dfrac{2x-2z+y}{y}+3\right)=\left(\dfrac{2y+2z-x}{x}+3\right)\)
\(\Leftrightarrow\dfrac{2x+2y+2z}{z}=\dfrac{2x+2y+2z}{y}=\dfrac{2x+2y+2z}{x}\)
\(\Leftrightarrow x=y=z\)
\(\Leftrightarrow M=\dfrac{2x.2y.2z}{8xyz}=1\)
Vậy \(\left[{}\begin{matrix}M=\dfrac{-1}{8}\Leftrightarrow x+y+z=0\\M=1\Leftrightarrow x+y+z\ne0\end{matrix}\right.\)
1.
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{x+y-3}{z}\)= \(\dfrac{2\left(x+y+z\right)}{x+y+z}=2\)
=> \(\dfrac{1}{x+y+z}\) = 2
=> x+y+z = \(\dfrac{1}{2}\)
Ta có: \(\dfrac{y+z+1}{x}\) = 2
=> y+z+1 = 2x => x+y+z+1 = 3x <=> \(\dfrac{3}{2}=3x\)
<=> x = \(\dfrac{1}{2}\)
Tương tự thế vào \(\dfrac{x+z+2}{y}\) tính được y =\(\dfrac{5}{6}\)
=> z = -\(\dfrac{5}{6}\)
=> A = 2016.\(\dfrac{1}{2}\) = 1008
Lời giải:
Từ đkđb suy ra:
$x-y=\frac{1}{z}-\frac{1}{y}=\frac{y-z}{yz}$
$y-z=\frac{1}{x}-\frac{1}{z}=\frac{z-x}{xz}$
$z-x=\frac{1}{y}-\frac{1}{x}=\frac{x-y}{xy}$
$\Rightarrow (x-y)(y-z)(z-x)=\frac{(y-z)(z-x)(x-y)}{(xyz)^2}$
$\Leftrightarrow (x-y)(y-z)(z-x)(1-\frac{1}{x^2y^2z^2})=0$
$\Rightarrow (x-y)(y-z)(z-x)=0$ hoặc $1-\frac{1}{x^2y^2z^2}=1$
$\Rightarrow (x-y)(y-z)(z-x)=0$ hoặc $x^2y^2z^2=1$
Nếu $(x-y)(y-z)(z-x)=0$
$\Rightarrow x=y$ hoặc $y=z$ hoặc $z=x$
Không mất tquat giả sử $x=y$. Khi đó: $\frac{1}{y}=\frac{1}{z}$
$\Rightarrow y=z$
$\Rightarrow x=y=z$. Tương tự khi xét $y=z$ hoặc $z=x$ thì ta cũng thu được $x=y=z$
Vậy $x=y=z$ hoặc $x^2y^2z^2=1$