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if-x-y-z-1-then-1-3xy-1-1-3yz-1-1-3zx-1-3-2xyz-




Question Number 147488 by mathdanisur last updated on 21/Jul/21
if  x;y;z≥1  then:  (1/(3xy−1)) + (1/(3yz−1)) + (1/(3zx−1)) ≥ (3/(2xyz))
$${if}\:\:{x};{y};{z}\geqslant\mathrm{1}\:\:{then}: \\ $$$$\frac{\mathrm{1}}{\mathrm{3}{xy}−\mathrm{1}}\:+\:\frac{\mathrm{1}}{\mathrm{3}{yz}−\mathrm{1}}\:+\:\frac{\mathrm{1}}{\mathrm{3}{zx}−\mathrm{1}}\:\geqslant\:\frac{\mathrm{3}}{\mathrm{2}{xyz}} \\ $$
Answered by mindispower last updated on 21/Jul/21
in firt  1≤xy,xy=((xyz)/z)  ⇒(1/(3xy−1))≥(1/(((3xyz)/z)−xy))=(z/(2xyz))  sam idea⇒(1/(3yz−1))≥(x/(2xyz)),(1/(3xz−1))≥(y/(2xyz))  LH≥(1/(2xyz))(x+y+z)  sinc x:y:z≥1  ⇒x+y+z≥3  ⇒(1/(3xy−1))+(1/(3zx−1))+(1/(3yz−1))≥((1.3)/(2xyz))=(3/(2xyz))
$${in}\:{firt} \\ $$$$\mathrm{1}\leqslant{xy},{xy}=\frac{{xyz}}{{z}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{3}{xy}−\mathrm{1}}\geqslant\frac{\mathrm{1}}{\frac{\mathrm{3}{xyz}}{{z}}−{xy}}=\frac{{z}}{\mathrm{2}{xyz}} \\ $$$${sam}\:{idea}\Rightarrow\frac{\mathrm{1}}{\mathrm{3}{yz}−\mathrm{1}}\geqslant\frac{{x}}{\mathrm{2}{xyz}},\frac{\mathrm{1}}{\mathrm{3}{xz}−\mathrm{1}}\geqslant\frac{{y}}{\mathrm{2}{xyz}} \\ $$$${LH}\geqslant\frac{\mathrm{1}}{\mathrm{2}{xyz}}\left({x}+{y}+{z}\right) \\ $$$${sinc}\:{x}:{y}:{z}\geqslant\mathrm{1} \\ $$$$\Rightarrow{x}+{y}+{z}\geqslant\mathrm{3} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{3}{xy}−\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{3}{zx}−\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{3}{yz}−\mathrm{1}}\geqslant\frac{\mathrm{1}.\mathrm{3}}{\mathrm{2}{xyz}}=\frac{\mathrm{3}}{\mathrm{2}{xyz}} \\ $$$$ \\ $$$$ \\ $$
Commented by mathdanisur last updated on 21/Jul/21
Thank you Ser cool
$${Thank}\:{you}\:{Ser}\:{cool} \\ $$

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