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Question-144684




Question Number 144684 by mnjuly1970 last updated on 27/Jun/21
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Answered by mindispower last updated on 27/Jun/21
M:=xyz−(xy+yz+zx)+x+y+z−1  (1/x)+(1/y)+(1/z)=1xyz=xy+yz+zx  M=x+y+z−1  harmonic arithmetic mean  ((x+y+z)/3)≥(3/((1/x)+(1/y)+(1/z)))=(3/1)=3  ⇒x+y+z≥9⇔x+y+z−1=M=8  equality x=y=z=3  (3−1)(3−1)(3−1)=8
$${M}:={xyz}−\left({xy}+{yz}+{zx}\right)+{x}+{y}+{z}−\mathrm{1} \\ $$$$\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}}=\mathrm{1}{xyz}={xy}+{yz}+{zx} \\ $$$${M}={x}+{y}+{z}−\mathrm{1} \\ $$$${harmonic}\:{arithmetic}\:{mean} \\ $$$$\frac{{x}+{y}+{z}}{\mathrm{3}}\geqslant\frac{\mathrm{3}}{\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}}}=\frac{\mathrm{3}}{\mathrm{1}}=\mathrm{3} \\ $$$$\Rightarrow{x}+{y}+{z}\geqslant\mathrm{9}\Leftrightarrow{x}+{y}+{z}−\mathrm{1}={M}=\mathrm{8} \\ $$$${equality}\:{x}={y}={z}=\mathrm{3} \\ $$$$\left(\mathrm{3}−\mathrm{1}\right)\left(\mathrm{3}−\mathrm{1}\right)\left(\mathrm{3}−\mathrm{1}\right)=\mathrm{8} \\ $$$$ \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 27/Jun/21
very nice mr power...
$${very}\:{nice}\:{mr}\:{power}… \\ $$

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