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Lời giải:
$x+y+z=2014; \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{2014}$
$\Rightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}$
$\Rightarrow (\frac{1}{x}+\frac{1}{y})+(\frac{1}{z}-\frac{1}{x+y+z})=0$
$\Rightarrow \frac{x+y}{xy}+\frac{x+y}{z(x+y+z)}=0$
$\Rightarrow (x+y)[\frac{1}{xy}+\frac{1}{z(x+y+z)}]=0$
$\Rightarrow (x+y).\frac{z(x+y+z)+xy}{xyz(x+y+z)}=0$
$\Rightarrow (x+y).\frac{(z+x)(z+y)}{xyz(x+y+z)}=0$
$\Rightarrow (x+y)(z+x)(z+y)=0$
$\Rightarrow x+y=0$ hoặc $x+z=0$ hoặc $z+y=0$
$\Rightarrow x=-y$ hoặc $y=-z$ hoặc $z=-x$
Vậy trong 3 số $x,y,z$ tồn tại hai số đối nhau.
\(\frac{2014x}{xy+2014x+2014}+\frac{y}{yz+y+2014}+\frac{z}{xz+z+1}=1\)
\(=\frac{xyz.x}{xy+xyzx+xyz}+\frac{y}{yz+y+xyz}+\frac{z}{zx+z+1}\)
\(=\frac{xz}{1+zx+z}+\frac{1}{z+1+zx}+\frac{z}{xz+z+1}=\frac{xz+1+z}{1+xz+z}=1\)
=> đpcm
Ta có:
\(M=\frac{x\left(yz-x^2\right)+y\left(zx-y^2\right)+z\left(xy-z^2\right)}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}=\frac{xyz-x^3+xyz-y^3+xyz-z^3}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}=\frac{3xyz-x^3-y^3-z^3}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}\)
\(-M=\frac{x^3+y^3+z^3-3xyz}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}\)
Xét đẳng thức phụ:
\(a^3+b^3+c^3-3abc=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=\left[\left(a +b\right)^3+c^3\right]-3ab\left(a+b+c\right)\)\(=\left(a+b+c\right)\left(\left(a+b\right)^2-c\left(a+b\right)+c^2\right)-ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2-ab\right]=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)\)
\(=\frac{1}{2}\left(a+b+c\right)\left(2a^2+2b^2+2c^2-2ab-abc-ac\right)\)
\(=\frac{1}{2}\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]\)
Thay vào -M ta có:
\(-M=\frac{\frac{1}{2}\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}=\frac{1}{2}\left(x+y+z\right)\Rightarrow M=-\frac{1}{2}\left(x+y+z\right)\)
Giờ thay: \(x=2014^{2015}-20142015;y=20142015-2015^{2014};z=2015^{2014}-2014^{2015}\)
Ta có:
\(M=-\frac{1}{2}\left(2014^{2015}-20142015+20142015-2015^{2014}+2015^{2014}-2014^{2015}\right)=0\)
Áp dụng BĐT AM-GM ta có:
\(x+y+z\ge3\sqrt[3]{xyz}\)
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge3\sqrt[3]{\frac{1}{xyz}}\)
Nhân theo vế 2 BĐT trên ta có:
\(VT\ge3^2\cdot\sqrt[3]{xyz\cdot\frac{1}{xyz}}=9=VP\)
Xảy ra khi \(a=b=c\)
Lời giải:
$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{2014}=\frac{1}{x+y+z}$
$\Leftrightarrow \frac{x+y}{xy}+\frac{1}{z}-\frac{1}{x+y+z}=0$
$\Leftrightarrow \frac{x+y}{xy}+\frac{x+y}{z(x+y+z)}=0$
$\Leftrightarrow (x+y)(\frac{1}{xy}+\frac{1}{z(x+y+z)})=0$
$\Leftrihgtarrow (x+y).\frac{z(x+y+z)+xy}{xyz(x+y+z)}=0$
$\Leftrightarrow (x+y).\frac{(z+x)(z+y)}{xyz(x+y+z)}=0$
$\Rightarrow (x+y)(y+z)(x+z)=0$
$\Leftrightarrow (2014-x)(2014-y)(2014-z)=0$
$\Leftrightarrow 2014-x=0$ hoặc $2014-y=0$ hoặc $2014-z=0$
$\Leftrightarrow x=2014$ hoặc $y=2014$ hoặc $z=2014$