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\(\left\{{}\begin{matrix}\left|x\right|\ge3\\\left|y\right|\ge3\\\left|z\right|\ge3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left|\dfrac{1}{x}\right|\le\dfrac{1}{3}\\\left|\dfrac{1}{y}\right|\le\dfrac{1}{3}\\\left|\dfrac{1}{z}\right|\le\dfrac{1}{3}\end{matrix}\right.\)
\(\left|A\right|=\left|\dfrac{xy+yz+xz}{xyz}\right|=\left|\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right|\le\left|\dfrac{1}{x}\right|+\left|\dfrac{1}{y}\right|+\left|\dfrac{1}{z}\right|\le\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{3}=1\)
\(\Rightarrow A\le\left|A\right|\le1\) (đpcm)
Dấu "=" xảy ra khi \(x=y=z=3\)
Ta có: \(x-y+z=0\)
\(\Rightarrow\left(x-y+z\right)^2=0
\)
\(\Rightarrow\left(x-y+z\right).\left(x-y+z\right)=0\)
\(\Rightarrow x\left(x-y+z\right)-y\left(x-y+z\right)+z\left(x-y+z\right)=0\)
\(\Rightarrow x^2-xy+xz-xy+y^2-yz+xz-yz+z^2=0\)
\(\Rightarrow x^2+y^2+z^2=xy+xy+yz+yz-xz-xz\)
\(\Rightarrow x^2+y^2+z^2=2xy+2yz-2xz\)
\(\Rightarrow x^2+y^2-z^2=2\left(xy+yz-xz\right)\)
Mà: \(x^2+y^2-z^2\ge0\)
\(\Rightarrow2\left(xy+yz-xz\right)\ge0\)
\(\Rightarrow xy+yz-xz\ge0\)(đpcm)
Vậy: \(xy+yz-xz\ge0\)
Áp dùng BĐT Cosi ta có:
\(\frac{x^3}{yz}+y+z\ge3\sqrt[3]{\frac{x^3}{yz}\cdot y\cdot z}=3x\)
\(\frac{y^3}{xz}+z+x\ge3\sqrt[3]{\frac{z^3}{zx}\cdot z\cdot x}=3y\)
\(\frac{z^3}{yx}+x+y\ge3\sqrt[3]{\frac{z^3}{xy}\cdot x\cdot y}=3z\)
\(\Rightarrow\frac{x^3}{xy}+y+z+\frac{y^3}{zx}+x+z+\frac{z^3}{xy}+x+y\ge3x+3y+3z\)
\(\Rightarrow\frac{x^3}{yz}+\frac{y^3}{xz}+\frac{z^3}{xy}\ge3\left(x+y+z\right)-2\left(x+y+z\right)\)\(=x+y+z\)
Dấu "=" xảy ra khi \(\hept{\begin{cases}\frac{x^3}{yz}=y=z\\\frac{y^3}{zx}=x=z\\\frac{z^3}{yz}=y=x\end{cases}\Rightarrow x=y=z}\)
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vì x - y - z = 0 nên x = y + z
Xét tổng A + B = xyz - xy2 - xz2 + y3 + z3
= ( y + z ) . yz - ( y + z ) . y2 - ( y + z ) . z2 + y3 + z3
= y2z + yz2 - y3 - y2z - yz2 - z3 + y3 + z3 = 0
Vậy ...
Chứng minh:
Ta có:
\(\left(x-y\right)^2\ge0\Rightarrow x^2+y^2-2xy\ge0\Rightarrow x^2+y^2\ge2xy\)
\(\left(y-z\right)^2\ge0\Rightarrow y^2+z^2-2yz\ge0\Rightarrow y^2+z^2\ge2yz\)
\(\left(x-z\right)^2\ge0\Rightarrow x^2+z^2-2xz\ge0\Rightarrow x^2+z^2\ge2xz\)
Cộng vế với vế, ta được:
\(2\left(x^2+y^2+z^2\right)\ge2\left(xy+yz+zx\right)\)
\(\Rightarrow3\left(x^2+y^2+z^2\right)\ge x^2+y^2+z^2+2\left(xy+yz+zx\right)\)
\(\Rightarrow x^2+y^2+z^2\ge\frac{1}{3}\left(x+y+z\right)^2\)(đpcm)
\(x^2+y^2+z^2\ge\frac{1}{3}\left(x+y+z\right)^2\)
\(\Leftrightarrow x^2+y^2+z^2-\frac{1}{3}\cdot\left(x+y+z\right)^2\ge0\)
\(\Leftrightarrow x^2+y^2+z^2-\frac{1}{3}\left(x^2+y^2+z^2+2xy+2yz+2xz\right)\ge0\)
\(\Leftrightarrow x^2+y^2+z^2-\frac{1}{3}\left(x^2+y^2+z^2\right)-\frac{2}{3}\left(xy+yz+zx\right)\ge0\)
\(\Leftrightarrow\frac{2}{3}\left(x^2+y^2+z^2\right)-\frac{2}{3}\left(xy+yz+xz\right)\ge0\)
\(\Leftrightarrow\frac{2}{3}\left(x^2+y^2+z^2-xy-yz-xz\right)\ge0\) (1)
Ta cần chứng minh : \(x^2+y^2+z^2-xy-yz-xz\ge0\)
\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2xz\ge0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2\ge0\) (luôn đúng)
=> bđt (1) đúng
\(\Rightarrow x^2+y^2+z^2\ge\frac{1}{3}\left(x+y+z\right)^2\) (đpcm)