$$frac1A=frac11 +frac11=frac1+11*1=frac21$$Hence $$A=1/2$$

Case 5$$frac1A=frac15 +frac15=frac5+55*5=frac1025$$Hence $$A=25/10=2.5$$

I can see in both of these cases $A$ is still $1$


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K-FeldsparK-Feldspar
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Since everything is strictly positive, it"s simply the fact that $$frac1A=frac1B+frac1C>frac1B$$ và $$frac1A>frac1Bimplies B>A$$ Same for $C$. Since $A$ is smaller than boh, it is smaller than the minimum.

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I don"t really understand why in you decided to lớn consider only three numbers.


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answered Nov 20, 2016 at 2:21
user228113user228113
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blackpen what trick? $endgroup$
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As you said, $frac1A=fracC+BC*B$which means $A=fracB*CB+C$Now clearly $B > fracB*CB+C$ because $fracB*CB+C = B*fracCB+C$ và $fracCB+C Similarly, $C > fracB*CB+C$ because $fracB*CB+C = C*fracBB+C$ and $fracBB+C Thus $B > fracB*CB+C = A$ & $C > fracB*CB+C = A$


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answered Nov 20, 2016 at 2:30
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