Subfield: Algebra


Question: Let $n$ be a positive integer, and let $x$ and $y$ be positive real numbers such that $x^{n}+y^{n}=1$. Prove that

$$
\left(\sum_{k=1}^{n} \frac{1+x^{2 k}}{1+x^{4 k}}\right)\left(\sum_{k=1}^{n} \frac{1+y^{2 k}}{1+y^{4 k}}\right)<\frac{1}{(1-x)(1-y)}
$$

Solution:
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

$$
\left(\sum_{k=1}^{n} \frac{1+x^{2 k}}{1+x^{4 k}}\right)\left(\sum_{k=1}^{n} \frac{1+y^{2 k}}{1+y^{4 k}}\right)<\frac{\left(\frac{1+\sqrt{2}}{2} \ln 2\right)^{2}}{(1-x)(1-y)}<\frac{0.7001}{(1-x)(1-y)}\tag{1}
$$

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

$$
f(t)=\frac{1+t}{1+t^{2}} \leq f\left(t_{0}\right)=\frac{1+\sqrt{2}}{2}=\alpha \text {. }
$$

Applying this to each term on the left-hand side of (1), we obtain

$$
\left(\sum_{k=1}^{n} \frac{1+x^{2 k}}{1+x^{4 k}}\right)\left(\sum_{k=1}^{n} \frac{1+y^{2 k}}{1+y^{4 k}}\right) \leq n \alpha \cdot n \alpha=(n \alpha)^{2} \tag{2}
$$

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

$$
(1-x)(1-y)=\exp g(t) \leq\left(1-2^{-1 / n}\right)^{2}\tag{3}
$$

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,

$$
\left(\sum_{k=1}^{n} \frac{1+x^{2 k}}{1+x^{4 k}}\right)\left(\sum_{k=1}^{n} \frac{1+y^{2 k}}{1+y^{4 k}}\right)<\frac{\left(\frac{1+\sqrt{2}}{2} \ln 2\right)^{2}}{(1-x)(1-y)}
$$