# Algebraic puzzle: Infer $x = \frac{y + z}{2}$ from $a^{\frac{1}{x}} = b^{\frac{1}{y}} + c^{\frac{1}{z}}$ and $a = b + c$

I have a suspicion that the following expression is true, however my algebra skills aren't brilliant, so any help would be appreciated:

Is it possible to infer $x = \frac{y + z}{2}$ from $a^{\frac{1}{x}} = b^{\frac{1}{y}} + c^{\frac{1}{z}}$ and $a = b + c$

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## 1 Answer

Let $a=2$, $b=c=1$. Also let $x=1$ then $a=b+c$ and $a^{\frac{1}{x}}=2$. Regardless of the values of $y$ and $z$ then $1^{\frac{1}{y}}+1^{\frac{1}{z}}=2$.

Thus we cannot infer that $x=\frac{y+z}{2}$

Counterexample: Let $a=2$, $b=c=1$, $x=1$ and $y=z=2$ then $b+c=2=a$ and $b^{\frac{1}{y}}+c^{\frac{1}{z}}=2=a^{\frac{1}{x}}$. But $x\neq \frac{y+z}{2}=2$.

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You may have misread the original problem. He's looking to infer that x = (y + z)/2, not (x + y)/2, and therefore your counterexample fails in the case where y = z = 1. – Alex Basson Sep 16 '10 at 10:26
But take $y=z=2$ then this is a counterexample. – alext87 Sep 16 '10 at 10:30