Prove that $\left (\sum_{k=1}^{n}\frac{1+x^{2k}}{1+x^{4k}} \right )\left ( \sum_{k=1}^{n}\frac{1+y^{2k}}{1+y^{4k}} \right )< \frac{1}{(1-x)(1-y)}.$ 
Let $n$ be a positive integer, and let $x$ and $y$ be positive real numbers such that $x^{n}+y^{n}=1$
Prove that $$\left (\sum_{k=1}^{n}\frac{1+x^{2k}}{1+x^{4k}}  \right )\left ( \sum_{k=1}^{n}\frac{1+y^{2k}}{1+y^{4k}} \right )< \frac{1}{(1-x)(1-y)}.$$

This is Olympiad question. What I have tried so far is...
we prove that $$\left (\sum_{k=1}^{n}\frac{1+x^{2k}}{1+x^{4k}}  \right )\left ( \sum_{k=1}^{n}\frac{1+y^{2k}}{1+y^{4k}} \right )<\frac{\left ( \frac{1+\sqrt{2}}{2} ln 2 \right )^{2}}{(1-x)(1-y)}< \frac{0.7001}{(1-x)(1-y)} $$
The idea is to estimate each term on the left-hand side with the same constant.
I don't know if my try is correct or not, also if it's correct I don't know how to complete, so any help will be appreciated, thanks.

For each real $t \in (0,1)$, $$\frac{1+t^{2}}{1+t^{4}}=\frac{1}{t}-\frac{(1-t)(1-t^{3})}{t(1+t^{4})}< \frac{1}{t}$$ Substituting $t=x^k$ and  $t=y^k$ $$0< \sum_{k=1}^{n}\frac{1+x^{2k}}{1+x^{4k}} < \sum_{k=0}^{n}\frac{1}{x^{k}}=\frac{1-x^{n}}{x^{n}(1-x)} \qquad and \qquad 0< \sum_{k=1}^{n}\frac{1+y^{2k}}{1+y^{4k}} < \sum_{k=0}^{n}\frac{1}{y^{k}}=\frac{1-y^{n}}{y^{n}(1-y)}$$ Since $1-y^n=x^n$ and $1-x^n=y^n$, $$\frac{1-x^{n}}{x^{n}(1-x)}=\frac{y^n}{x^{n}(1-x)}, \qquad \frac{1-y^{n}}{y^{n}(1-y)}=\frac{x^n}{y^{n}(1-y)}$$
and therefore $$\left (\sum_{k=1}^{n}\frac{1+x^{2k}}{1+x^{4k}}  \right )\left ( \sum_{k=1}^{n}\frac{1+y^{2k}}{1+y^{4k}} \right )< \frac{y^n}{x^{n}(1-x)} \cdot \frac{x^n}{y^{n}(1-y)}=\frac{1}{(1-x)(1-y)}$$
AND WE'RE DONE
