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I want to prove the proposition in the title, if it is at all possible. From computational evidence, I have a hunch that it is true. How would I go about this? If this is too strong a conjecture, then I would be satisfied to know how to do it for $x=e^2$. Many thanks in advance.

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HINT: $x^y\cdot x^{-y}=x^0=1$, so it suffices to show that if $0<y<1$, and $xy=1$, then $x+y\ge 2$. Consider $(x+y)^2$.

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Call $t=x^y>0$. What is the minimum of the function $f(t)=t+\frac1t$?

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If $z>0$ then $z+1/z\geq 2$ because $z+1/z-2=(\sqrt z -1/\sqrt z)^2.$

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