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Question Number 115951 by Fikret last updated on 29/Sep/20

$$x,y,zϵR^{+}$$$$2x+3y+4z=1\Rightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}smallestintegervalue?$$

Answered by 1549442205PVT last updated on 30/Sep/20

$$\mathrm{From}\mathrm{the}\mathrm{hypothesis}\mathrm{we}\mathrm{have}$$$$\mathrm{P}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{2x+3y+4z}{\mathrm{x}}+\frac{2x+3y+4z}{\mathrm{y}}$$$$+\frac{2x+3y+4z}{\mathrm{z}}=2+3+4+3\frac{\mathrm{y}}{\mathrm{x}}+2\frac{\mathrm{x}}{\mathrm{y}}$$$$+4\frac{\mathrm{z}}{\mathrm{x}}+2\frac{\mathrm{x}}{\mathrm{z}}+3\frac{\mathrm{y}}{\mathrm{z}}+4\frac{\mathrm{z}}{\mathrm{y}}$$$$⩾2+3+4+3+2+4+2+3+4=27$$$$(\mathrm{Since}\mathrm{x},\mathrm{yz}>0,\frac{\mathrm{x}}{\mathrm{y}},\frac{\mathrm{y}}{\mathrm{x}},\frac{\mathrm{y}}{\mathrm{z}},\frac{\mathrm{z}}{\mathrm{y}},\frac{\mathrm{x}}{\mathrm{z}},\frac{\mathrm{z}}{\mathrm{x}}\mathrm{has}$$$$\mathrm{least}\mathrm{integeral}\mathrm{value}\mathrm{equal}\mathrm{to}1)$$$$\mathrm{The}\mathrm{equality}\mathrm{ocurrs}\mathrm{if}\mathrm{and}\mathrm{only}\mathrm{if}$$$$\{\begin{array}{l}2\mathrm{x}+3\mathrm{y}+4\mathrm{z}=1\\ \mathrm{x}=\mathrm{y}=\mathrm{z}\end{array}\iff \mathrm{x}=\mathrm{y}=\mathrm{z}=\frac{1}{9}$$$$\mathrm{Thus},\mathrm{the}\mathrm{least}\mathrm{integral}\mathrm{value}\mathrm{of}\mathrm{P}\mathrm{is}$$$${(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})}_{\min }=27\mathrm{when}$$$$\mathrm{x}=\mathrm{y}=\mathrm{z}=1/9$$

Commented by soumyasaha last updated on 01/Oct/20

$$$$$$\mathrm{Assuming}\mathrm{x},\mathrm{y},\mathrm{z}\mathrm{positive},\mathrm{we}\mathrm{have}$$$$\frac{2\mathrm{x}+3\mathrm{y}+4\mathrm{z}}{3}⩾\sqrt[3]{2\mathrm{x}.3\mathrm{y}.4\mathrm{z}}$$$$\Rightarrow 24\mathrm{xyz}⩽\frac{1}{27}$$$$\Rightarrow \frac{1}{\mathrm{xyz}}⩾27.24..........\left(\mathrm{i}\right)$$$$$$$$\mathrm{Again},\frac{\frac{1}{\mathrm{x}}+\frac{1}{\mathrm{y}}+\frac{1}{\mathrm{z}}}{3}⩾\sqrt[3]{\frac{1}{\mathrm{x}}.\frac{1}{\mathrm{y}}.\frac{1}{\mathrm{y}}}$$$$\Rightarrow \frac{1}{\mathrm{x}}+\frac{1}{\mathrm{y}}+\frac{1}{\mathrm{z}}⩾3.\sqrt[3]{\frac{1}{\mathrm{xyz}}}$$$$\Rightarrow \frac{1}{\mathrm{x}}+\frac{1}{\mathrm{y}}+\frac{1}{\mathrm{z}}⩾3.\sqrt[3]{27.24}$$$$\Rightarrow \frac{1}{\mathrm{x}}+\frac{1}{\mathrm{y}}+\frac{1}{\mathrm{z}}⩾18.\sqrt[3]{3}$$$$\therefore \mathrm{least}\mathrm{integral}\mathrm{value}\mathrm{is}26$$

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