Prove that for all real numbers $x,y,z$:
$$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\implies \frac{1}{x^5}+\frac{1}{y^5}+\frac{1}{z^5}=\frac{1}{(x+y+z)^5}.$$
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Michael Rozenberg
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Mel
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are $x,y,z$ positive reals? – Dr. Sonnhard Graubner Sep 04 '17 at 18:10
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i would think we can use $AM-GM$ if the variables are positive – Dr. Sonnhard Graubner Sep 04 '17 at 18:12
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The condition gives $$(x+y+z)(xy+xz+yz)=xyz$$ or $$(x+y)(x+z)(y+z)=0$$ and since for $y=-x$ it's true and we need to prove something symmetry, we are done!
Michael Rozenberg
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Hint: We have $$ 0=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-\frac{1}{x+y+z}=\frac{(x + y)(x + z)(y + z)}{(y + z + x)xyz}. $$ Hence we have $$(x+y)(x+z)(y+z)=0.$$
Dietrich Burde
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