My first instinct was to show that the derivative is always negative for the conditions above and with the limit $=0$ it should follow that it is always positive but it is way to complicated for me. This can be done for every $d$ I consider. I can't find a more direct way.
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$\begingroup$ Hint: Both terms are positive, so you can rearrange and take the inverse, and then you just have to show that $\left(\dfrac{2x+2}{d}\right)^d>x$. $\endgroup$– TonyKCommented Oct 5, 2020 at 14:39
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$\begingroup$ Does it help to consider $\left(\dfrac{2x+2}{d}\right)^d-x$ as polynomial and search for roots? It might be possible to do it for each $d$ differently $\endgroup$– LocusCommented Oct 5, 2020 at 14:53
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1 Answer
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When the cases are small we can check each case separately. So first substitute d=2 in the function. $$f(x)=\frac1x-(\frac{2}{2x+2})^2$$ Now find the interval of $x$ for which $f(x)>0$ by solving the inequality: $$\frac1x-(\frac{2}{2x+2})^2>0$$ $$\frac1x>(\frac{2}{2x+2})^2$$ $$\frac1x>\frac{4}{(2x+2)^2}$$ $$\frac1x>\frac{4}{(2x+2)^2}$$ $$(2x+2)^2>4x$$ $$4x^2+4+8x>4x$$ $$4x^2+4+4x>0$$ $$x^2+1+x>0$$ Which is greater then zero for all $x\epsilon (-\infty,\infty)$, Similarly you can check cases for $d=3,4,5$.
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$\begingroup$ I don't think the cases for $x>2$ will be so simple! $\endgroup$– TonyKCommented Oct 5, 2020 at 17:47
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$\begingroup$ yes those would be more complex. $\endgroup$ Commented Oct 6, 2020 at 18:03