IMO Shortlist, 2007

For each real $t \in (0,1)$,
$$\frac{1+t^{2}}{1+t^{4}}=\frac{1}{t}-\frac{(1-t)\left(1-t^{3}\right)}{t\left(1+t^{4}\right)}<\frac{1}{t}$$
Substituting $t=x^{k}$ and $t=y^{k}$,
$$0<\sum_{k=1}^{n} \frac{1+x^{2 k}}{1+x^{4 k}}<\sum_{k=1}^{n} \frac{1}{x^{k}}=\frac{1-x^{n}}{x^{n}(1-x)} \quad \text{ and } \quad 0<\sum_{k=1}^{n} \frac{1+y^{2 k}}{1+y^{4 k}}<\sum_{k=1}^{n} \frac{1}{y^{k}}=\frac{1-y^{n}}{y^{n}(1-y)} .$$
Since $1-y^{n}=x^{n}$ and $1-x^{n}=y^{n}$,
$$\frac{1-x^{n}}{x^{n}(1-x)}=\frac{y^{n}}{x^{n}(1-x)}, \quad \frac{1-y^{n}}{y^{n}(1-y)}=\frac{x^{n}}{y^{n}(1-y)}$$
and therefore
$$\left(\sum_{k=1}^{n} \frac{1+x^{2 k}}{1+x^{4 k}}\right)\left(\sum_{k=1}^{n} \frac{1+y^{2 k}}{1+y^{4 k}}\right)<\frac{y^{n}}{x^{n}(1-x)} \cdot \frac{x^{n}}{y^{n}(1-y)}=\frac{1}{(1-x)(1-y)} .$$

We prove
$$\begin{equation*} \left(\sum_{k=1}^{n} \frac{1+x^{2 k}}{1+x^{4 k}}\right)\left(\sum_{k=1}^{n} \frac{1+y^{2 k}}{1+y^{4 k}}\right)<\frac{\left(\frac{1+\sqrt{2}}{2} \ln 2\right)^{2}}{(1-x)(1-y)}<\frac{0.7001}{(1-x)(1-y)} \tag{1} \end{equation*}$$
The idea is to estimate each term on the left-hand side with the same constant. To find the upper bound for the expression $\frac{1+x^{2 k}}{1+x^{4 k}}$, consider the function $f(t)=\frac{1+t}{1+t^{2}}$ in interval $(0,1)$. Since
$$f^{\prime}(t)=\frac{1-2 t-t^{2}}{\left(1+t^{2}\right)^{2}}=\frac{(\sqrt{2}+1+t)(\sqrt{2}-1-t)}{\left(1+t^{2}\right)^{2}}$$
the function increases in interval $(0, \sqrt{2}-1]$ and decreases in $[\sqrt{2}-1,1)$. Therefore the maximum is at point $t_{0}=\sqrt{2}-1$ and
$$f(t)=\frac{1+t}{1+t^{2}} \leq f\left(t_{0}\right)=\frac{1+\sqrt{2}}{2}=\alpha$$
Applying this to each term on the left-hand side of (1), we obtain
$$\begin{equation*} \left(\sum_{k=1}^{n} \frac{1+x^{2 k}}{1+x^{4 k}}\right)\left(\sum_{k=1}^{n} \frac{1+y^{2 k}}{1+y^{4 k}}\right) \leq n \alpha \cdot n \alpha=(n \alpha)^{2} . \tag{2} \end{equation*}$$
To estimate $(1-x)(1-y)$ on the right-hand side, consider the function
$$g(t)=\ln \left(1-t^{1 / n}\right)+\ln \left(1-(1-t)^{1 / n}\right)$$
Substituting $s$ for $1-t$, we have
$$-n g^{\prime}(t)=\frac{t^{1 / n-1}}{1-t^{1 / n}}-\frac{s^{1 / n-1}}{1-s^{1 / n}}=\frac{1}{s t}\left(\frac{(1-t) t^{1 / n}}{1-t^{1 / n}}-\frac{(1-s) s^{1 / n}}{1-s^{1 / n}}\right)=\frac{h(t)-h(s)}{s t} .$$
The function
$$h(t)=t^{1 / n} \frac{1-t}{1-t^{1 / n}}=\sum_{i=1}^{n} t^{i / n}$$
is obviously increasing for $t \in (0,1)$, hence for these values of $t$ we have
$$g^{\prime}(t)>0 \Longleftrightarrow h(t)<h(s) \Longleftrightarrow t<s=1-t \Longleftrightarrow t<\frac{1}{2} .$$
Then, the maximum of $g(t)$ in ( 0,1 ) is attained at point $t_{1}=1 / 2$ and therefore
$$g(t) \leq g\left(\frac{1}{2}\right)=2 \ln \left(1-2^{-1 / n}\right), \quad t \in (0,1)$$
Substituting $t=x^{n}$, we have $1-t=y^{n},(1-x)(1-y)=\exp g(t)$ and hence
$$\begin{equation*} (1-x)(1-y)=\exp g(t) \leq\left(1-2^{-1 / n}\right)^{2} \tag{3} \end{equation*}$$
Combining (2) and (3), we get
$$\left(\sum_{k=1}^{n} \frac{1+x^{2 k}}{1+x^{4 k}}\right)\left(\sum_{k=1}^{n} \frac{1+y^{2 k}}{1+y^{4 k}}\right) \leq(\alpha n)^{2} \cdot 1 \leq(\alpha n)^{2} \frac{\left(1-2^{-1 / n}\right)^{2}}{(1-x)(1-y)}=\frac{\left(\alpha n\left(1-2^{-1 / n}\right)\right)^{2}}{(1-x)(1-y)} .$$
Applying the inequality $1-\exp (-t)<t$ for $t=\frac{\ln 2}{n}$, we obtain
$$\alpha n\left(1-2^{-1 / n}\right)=\alpha n\left(1-\exp \left(-\frac{\ln 2}{n}\right)\right)<\alpha n \cdot \frac{\ln 2}{n}=\alpha \ln 2=\frac{1+\sqrt{2}}{2} \ln 2 .$$
Hence,
$$\left(\sum_{k=1}^{n} \frac{1+x^{2 k}}{1+x^{4 k}}\right)\left(\sum_{k=1}^{n} \frac{1+y^{2 k}}{1+y^{4 k}}\right)<\frac{\left(\frac{1+\sqrt{2}}{2} \ln 2\right)^{2}}{(1-x)(1-y)}$$