# A3

first note that we have x,y\in\mathbb{R^+}\implies x^n,y^n\in\mathbb{R^+} for some n\in\mathbb{Z^+}
so from this it is obvious that, using the fact x^n+y^n=1, x^n>0\iff1-y^n>0\iff0<y^n<1\iff0<y<1
and similarly y^n>0\iff1-x^n>0\iff0<x^n<1\iff0<x<1.
this means that \dfrac{1}{1-x}>0 and \dfrac{1}{1-y}>0 and taking their product won’t invalidate any inequality results.

due to the above, the inequality is equivalent to proving, that both

\sum_{k=1}^{n}\frac{1+x^{2k}}{1+x^{4k}}<\frac{1}{1-x}

and

\sum_{k=1}^{n}\frac{1+y^{2k}}{1+y^{4k}}<\frac{1}{1-y}

are true, separately (then we can take their product).
due to the similarity of the two inequalities, we can prove one and the other will follow.

we need to somehow bound the sum by above.
let us manipulate \displaystyle \sum_{k=1}^{n}\frac{1+y^{2k}}{1+y^{4k}}, WLOG.

\begin{align} \sum_{k=1}^{n}\frac{1+y^{2k}}{1+y^{4k}} &=\sum_{k=1}^{n}\dfrac{y^k(1+y^{2k})}{y^k(1+y^{4k})} \\&=-\sum_{k=1}^{n}\dfrac{-y^k-y^{3k}}{y^k(1+y^{4k})} \\&=-\sum_{k=1}^{n}\dfrac{1-y^k-y^{3k}+y^{4k}-(1+y^{4k})}{y^k(1+y^{4k})} \\&=\sum_{k=1}^{n}\dfrac{(1+y^{4k})-(1-y^k)(1-y^{3k})}{y^k(1+y^{4k})} \\&=\sum_{k=1}^{n}\left(\dfrac{1}{y^k}-\dfrac{(1-y^k)(1-y^{3k})}{y^k(1+y^{4k})}\right) \end{align}

as 0<y<1\implies 0<y^k<1 it is clear that \dfrac{(1-y^k)(1-y^{3k})}{y^k(1+y^{4k})}\in\mathbb{R^+} and so we get that

\begin{align} \sum_{k=1}^{n}\left(\dfrac{1}{y^k}-\dfrac{(1-y^k)(1-y^{3k})}{y^k(1+y^{4k})}\right) &< \sum_{k=1}^{n}\dfrac{1}{y^k} \\&= \frac{y^{-1}(y^{-n}-1)}{y^{-1}-1} \\&= \frac{y^{-n}-1}{1-y} \\&= \frac{1-y^n}{y^n(1-y)} \end{align}

by the exact same argument we would get

\sum_{k=1}^{n}\frac{1+x^{2k}}{1+x^{4k}}<\frac{1-x^n}{x^n(1-x)}
\begin{align} \therefore\left(\sum_{k=1}^{n}\frac{1+x^{2k}}{1+x^{4k}}\right)\left(\sum_{k=1}^{n}\frac{1+y^{2k}}{1+y^{4k}}\right) &<\left(\frac{1-x^n}{x^n(1-x)}\right)\left(\frac{1-y^n}{y^n(1-y)}\right) \\&= \left(\frac{\cancel{y^n}}{\cancel{x^n}(1-x)}\right)\left(\frac{\cancel{x^n}}{\cancel{y^n}(1-y)}\right) \\ &= \frac{1}{(1-x)(1-y)} \end{align}

as needed, due to the relation x^n+y^n=1.

can kind of see how they created the question (still took me trying out like 7 different things to get the result), satisfyingly cancelling like that can’t be a coincidence

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