# For $z =\frac{ xy}{x+y}$, why is $z < x$ and $z < y$ for all values of $x > 0$ and $y > 0$?

For $z =\frac{ xy}{x+y}$, why is $z < x$ and $z < y$ for all values of $x > 0$ and $y > 0$?

This question has to do with the concept of resistors, given two resistors in parallel the equivalent resistance is always lower than the smallest individual resistance, I am trying to convince myself that this is true.

The equivalent resistance is given by $z$ and the individual resistances are $x$ and $y$.

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Please use parentheses. $xy/x+y=(xy/x)+y$ which is not what you meant. – Ross Millikan Sep 30 '12 at 22:37
suppose you write it $$\frac{1}{z} = \frac{1}{x} + \frac{1}{y}$$ and try to show $$\frac{1}{z} > \frac{1}{x}$$ and so on... – GEdgar Sep 30 '12 at 22:41

Note that $$z={xy\over x+y} = x\cdot {y\over x+y}$$ and that ${y \over x+y}<1$ since the numerator is smaller than the denominator.

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Jeez such a simple explanation, can't believe I did not realize it. Thanks, will choose your answer as soon as it lets me. – user1084113 Sep 30 '12 at 22:46

Because if $0<a<b$ then $\displaystyle\frac ca>\frac cb$. So, $$\frac{xy}{x+y}<\frac{xy}x$$

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$x,y>0$ implies $$z=\frac{xy}{x+y}=\frac x{x+y}\cdot y<\frac{x+y}{x+y}\cdot y=y$$ and the same with $x$, $y$ interchanged.

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I feel like I am thinking too complicated sometimes. I had proven it like that:

$$z=\frac{xy}{x+y}=\frac{xy}{x\cdot (1+\frac{y}{x})}=\frac{y}{1+\frac{y}{x}}$$

With $x,y>0$ follows

$$1+\frac{y}{x}>1\rightarrow\frac{y}{1+\frac{y}{x}}<y$$

This makes $z<y$ and $z<x$ respectively.

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