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If
\[\huge {1\over z} = {1 \over x } + {1 \over y}\]
and x>y>0
prove that z
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if \[\huge {1\over z }= {1 \over x} + {1 \over y} \] and x>y>0 prove that z<y
Since \(x,y>0\), we know that \(1/x+1/y>1/y\). So \(1/z>1/y\), so \(z>y\).
another way to prove this is 1/z = 1/x + 1/y = (x+y)/(xy) then z = (xy)/(x+y) since x,y > 0, (xy)/(x+y) < (xy)/x = y so z < y
thank you guys :)
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