By Cauchy–Schwarz inequality: $(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}) \geq (1+1+1)^2 = 9$ By condition: $(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}) = 2010 \cdot \frac{3}{670} = 9$ Therefore, we have equality, and thats iff $\frac{x}{\frac{1}{x}} = \frac{y}{\frac{1}{y}} = \frac{z}{\frac{1}{z}}$ , so $x^2=y^2=z^2$, and (because of ${R}_{+}^{*}$) $x=y=z$. Therefore: $x=y=z=670$.

alternate: AM Of $\frac{x+y+z}{3}=\frac{1}{670}$ , HMOf $\frac{3}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}=\frac{1}{670}$ AM=HM $\implies{x=y=z=\frac{1}{670}}$

momo1729 wrote: Solve the following equation in $\mathbb{R}^+$ : \[\left\{\begin{matrix} \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2010\\ x+y+z=\frac{3}{670} \end{matrix}\right.\] x=y=z=1/670

Using AM- $\frac{x+y+z}{3}=\frac{1}{670}$ so $x=y=z$ --> $3x=\frac{3}{670}$ ----> $x=\frac{1}{670}$

x=y=z=1\670 only

master_Hjom wrote: By Cauchy–Schwarz inequality: $(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}) \geq (1+1+1)^2 = 9$ By condition: $(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}) = 2010 \cdot \frac{3}{670} = 9$ Therefore, we have equality, and thats iff $\frac{x}{\frac{1}{x}} = \frac{y}{\frac{1}{y}} = \frac{z}{\frac{1}{z}}$ , so $x^2=y^2=z^2$, and (because of ${R}_{+}^{*}$) $x=y=z$. Therefore: $x=y=z=670$. $x=y=z=(670)^{–1}$

$z=x+y+z\ge 3(xyz)^{\frac{1}{3}}$...................(1) $p=\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \ge 3(xyz)^{\frac{–1}{3}}$...........(2) $pz\ge 9$ equality at $x=y=z$ Our original equation becomes $3x=\frac{3}{670}$ ....and so it follows Relatively easy .....