We have: $ \frac {1 + x^{2k}}{1 + x^{4k}} < x^{ - 2k}$. Then we can write: $ \sum^n_{k = 1} \frac {1 + x^{2k}}{1 + x^{4k}} < \sum^n_{k = 1} x^{ - 2k} = \frac {1}{1 - x^2}\cdot (x^{ - 2n} - 1) = \frac {1}{1 - x^2}\cdot \frac {1 - x^{2n}}{x^{2n}} = \frac {1}{1 - x^2}\cdot \frac {(1 - x^{n})(1 + x^n)}{x^{2n}} = \frac {1}{1 - x^2}\cdot \frac {y^{n}(1 + x^n)}{x^{2n}}$ And: $ \left(\sum^n_{k = 1} \frac {1 + x^{2k}}{1 + x^{4k}} \right) \cdot \left( \sum^n_{k = 1} \frac {1 + y^{2k}}{1 + y^{4k}} \right) < \frac {1}{1 - x^2}\cdot \frac {y^{n}(1 + x^n)}{x^{2n}}\cdot \frac {1}{1 - y^2}\cdot \frac {x^{n}(1 + y^n)}{y^{2n}} = \frac {1}{1 - x^2}\cdot \frac {1 + x^n}{x^{n}}\cdot \frac {1}{1 - y^2}\cdot \frac {1 + y^n}{y^{n}} = \frac {1}{1 - x^2}\cdot \frac {1 + x^n}{1 - y^{n}}\cdot \frac {1}{1 - y^2}\cdot \frac {1 + y^n}{1 - x^{n}} = \frac {1}{1 - x^2}\cdot \frac {1 + x^n}{1 - x^{n}}\cdot \frac {1}{1 - y^2}\cdot \frac {1 + y^n}{1 - y^{n}} < \frac {1}{(1 - x) \cdot (1 - y)}.$ This gives: $ (1 + x^n)(1 + y^n) < (1 + x)(1 + y)(1 - x^n)(1 - y^n)$ or $ 1 + x^n + y^n + x^ny^n < (1 + x + y + xy)(1 - x^n - y^n + x^ny^n)$ or $ 2 + x^ny^n < (1 + x + y + xy)x^ny^n$ $ 2 + x^ny^n < x^ny^n + x^ny^n (x + y + xy)$ $ 2 < x^ny^n (x + y + xy)$ This is not true If $ y \to 0$ we get false. Think there is some condition $ \frac {1 + x^{2k}}{1 + x^{4k}} < ?$ but I don't know what.

really nice solution . if y near to zero the right hand side is very great and hold that inequality , am i right?

My first case $ \frac {1 + x^{2k}}{1 + x^{4k}} < x^{ - 2k}$ does not give the result at the end. Inequality does not hold with it.

Since $ t \mapsto x^t$ is a strictly convex function, by Karamata's inequality, we have \begin{align*} x^k + x^{k+4} + x^{3k+2} + x^{3k+6} &< x^0 + x^2 + x^{4k+4} + x^{4k+6} \\ x^k(1+x^4)(1+x^{2k+2}) &< (1+x^2)(1+x^{4k+4}) \\ \frac{1+x^{2k+2}}{1+x^{4k+4}} &< \frac{1+x^2}{1+x^4} \cdot \frac{1}{x^k} \end{align*} for any nonnegative integer $ k$, since the sequence $ (4k+6,4k+4,2,0)$ strictly majorizes the sequence $ (3k+6,3k+2,k+4,k)$. (One can also prove this inequality by repeated application of weighted AM-GM.) Thus \begin{align*} \sum_{k=1}^n \frac{1+x^{2k}}{1+x^{4k}} &< \sum_{k=0}^{n-1} \frac{1+x^2}{1+x^4} \cdot \frac{1}{x^k} \\ &= \frac{1+x^2}{1+x^4} \cdot \frac{1/x^n - 1}{1/x-1} \\ &= \frac{(1+x^2)x(1-x^n)}{(1+x^4)x^n(1-x)} . \end{align*} Again by Karamata, $ (1+x^2)x = x+x^3 \le 1+x^4$, so \[ \frac{(1+x^2)x(1-x^n)}{(1+x^4)x^n(1-x)} \le \frac{(1-x^n)}{x^n(1-x)} = \frac{y^n}{x^n(1-x)} , \] since $ x^n + y^n = 1$. It now follows that \[ \biggl( \sum_{k=1}^n \frac{1+x^{2k}}{1+x^{4k}} \biggr) \cdot \biggl( \sum_{k=1}^n \frac{1+y^{2k}}{1+y^{4k}} \biggr) < \frac{y^n}{x^n(1-x)} \cdot \frac{x^n}{y^n(1-y)} = \frac{1}{(1-x)(1-y)}, \] as desired, since all quanties in question are positive. $ \blacksquare$

Nice solution, Boy Soprano II.

A bit late, but- Solution) $ \frac{1+x^{2k}}{1+x^{4k}} < \frac{1}{x}$ (Equivalent to $ x^4 - x^3 - x + 1 > 0$ - equality holds only for $ x=1$ which is impossible) Therefore, $ LHS \leq \left(\sum^{n}_{k=1}\frac{1}{x}\right)\cdot\left(\sum^{n}_{k=1}\frac{1}{y}\right)$ $ = \frac{1-x^n}{(1-x)x^n} \cdot \frac{1-y^n}{(1-y)y^n} = \frac{y^n}{(1-x)x^n} \cdot \frac{x^n}{(1-y)y^n} = RHS$

orl wrote: Let $ n$ be a positive integer, and let $ x$ and $ y$ be a positive real number such that $ x^n + y^n = 1.$ Prove that \[ \left(\sum^n_{k = 1} \frac {1 + x^{2k}}{1 + x^{4k}} \right) \cdot \left( \sum^n_{k = 1} \frac {1 + y^{2k}}{1 + y^{4k}} \right) < \frac {1}{(1 - x) \cdot (1 - y)}.\] Author: unknown author, Estonia $ \frac{1+x^{2k}}{1+x^{4k}} < \frac{1}{x^k}$. So: $ \sum^n_{k = 1} \frac {1 + x^{2k}}{1 + x^{4k}} < \frac{\left ( \frac{1}{x^n} \right ) - 1}{1-x} = \frac{1}{1-x} \cdot \frac{y^n}{x^n}$ So $ LHS < \frac{1}{1-x} \cdot \frac{y^n}{x^n} \cdot \frac{1}{1-y} \cdot \frac{x^n}{y^n} = RHS$

Since $x,y>0$ and $x^n + y^n = 1$, it follows that $x,y \in (0,1)$. Hence for any $k \in \mathbb{N}$, it follows that \[(1-x^{3k})(1-x^{k})>0\] Rearranging this inequality yields that \[\frac{1+x^{2k}}{1+x^{4k}} < \frac{1}{x^k}\] Summing these inequalities for $k=1,2,\dots, n$ yields that \[\sum^{n}_{k = 1}\frac{1+x^{2k}}{1+x^{4k}}<\sum^{n}_{k = 1}\frac{1}{x^k}=\frac{1}{x^n}\sum^{n-1}_{k = 0}x^k = \frac{1}{x^n} \cdot \frac{1-x^n}{1-x}=\frac{y^n}{x^n (1-x)}\] Similarly, it follows that \[\sum^{n}_{k = 1}\frac{1+y^{2k}}{1+y^{4k}}<\frac{x^n}{y^n (1-y)}\] Multiplying these two inequalities yields that \[ \left(\sum^{n}_{k = 1}\frac{1+x^{2k}}{1+x^{4k}}\right)\cdot\left(\sum^{n}_{k = 1}\frac{1+y^{2k}}{1+y^{4k}}\right) < \frac{1}{(1-x)\cdot (1-y)} \]

This problem is nice. First of all we have $\frac{1+x^{2k}}{1+x^{4k}} < \frac{1+x^{2k}}{\frac{(1+x^{2k})^2}{2}}=\frac{2}{1+x^{2k}} < \frac{2}{2x^k}=\frac{1}{x^k}$ So summing over we get $\sum_{k=1}^{n}{\frac{1+x^{2k}}{1+x^{4k}} < \sum_{k=1}^{n}{\frac{1}{x^k}}=\frac{(1-x^n)}{x^n(1-x)}=\frac{y^n}{x^n(1-x)}}$ Similarly $\sum_{k=1}^{n}{\frac{1+y^{2k}}{1+y^{4k}} < \frac{x^n}{y^n(1-y)}}$ Multiplying these inequalities we obtain the result. So we are done.

Note that $x^k<1\implies (1-x^k)(1-x^{3k})>0\implies \frac{1+x^{2k}}{1+x^{4k}}<\frac{1}{x^k}$. So, $$LHS<\left(\sum_{k=1}^n\frac{1}{x^k}\right)\left(\sum_{k=1}^n\frac{1}{y^k}\right)=\frac{1-x^n}{x^n(1-x)}\frac{1-y^n}{y^n(1-y)}=\frac{1}{(1-x)(1-y)}$$

The crux of the proof is the following bound: \begin{align*} \frac{1 + x^{2k}}{1 + x^{4k}} < \frac{1}{x^k}. \end{align*}It's equivalent to prove \begin{align*} 0 < x^4 - x^3 - x + 1 = (x - 1)^2(x^2 + x + 1), \end{align*}which is true for $x$ over the interval $(0, 1)$. Using the bound yields \begin{align*} \text{LHS} &< \left(\frac 1 x + \frac{1}{x^2} + \cdots + \frac{1}{x^n} \right) \cdot \left(\frac 1 y + \frac{1}{y^2} + \cdots + \frac{1}{y^n} \right) \\ &= \frac{1}{x^n} \cdot \frac{1}{y^n} \cdot \frac{(1 - x^n)(1 - y^n)}{(1 - x)(1 - y)} \\ &= \frac{1}{(1 - x)(1 - y)}, \end{align*}as desired.

Solved with Sross314. Claim: $\frac{1+x^{2k}}{1+x^{4k}}<\frac{1}{x^k}$ for $0<x<1$ and positive integers $k$. Proof: This inequality is equivalent to \[x^k(x^{2k}+1)<x^{4k}+1,\]which is equivalent to \[x^{4k}-x^{3k}-x^k+1>0.\] Now we have \[x^{4k}-x^{3k}-x^k+1=(x^k-1)^2(x^{2k}+x^k+1)>0\]$\blacksquare$ It suffices to show that \[\left(\frac{1}{x}+\frac{1}{x^2}+\cdots+\frac{1}{x^n}\right)\left(\frac{1}{y}+\frac{1}{y^2}+\cdots+\frac{1}{y^n}\right)\le \frac{1}{(1-x)(1-y)} \] We have \[\sum_{k=1}^n \frac{1}{x^k}=\frac{x^{n-1}+x^{n-2}+\cdots+x+1}{x^n}=\frac{\tfrac{x^{n}-1}{x-1}}{x^n}=\frac{x^{n}-1}{x^n(x-1)}\] Multiplying by $\frac{y^n-1}{y^n(y-1)}$ gives \[\frac{(1-x^n)(1-y^n)}{x^ny^n(1-x)(1-y)}=\frac{1}{(1-x)(1-y)},\]so we are done.

Notice that $\forall x^k \in (0,1)$ $$x^{4k}-x^{3k}-x^{k}+1=(1-x^k)(1-x^{3k})>0 \iff \frac{1}{x^k} > \frac{1+x^{2k}}{1+x^{4k}}$$Therefore $$\frac{1}{(1-x)(1-y)}=\sum_{k=1}^n \frac{1}{x^k} \sum_{k=1}^n\frac{1}{y^k} >\left (\sum_{k=1}^n \frac{1+x^{2k}}{1+x^{4k}}\right )\left ( \sum_{k=1}^n \frac{1+y^{2k}}{1+y^{4k}}\right)$$

Boy Soprano II wrote: Since $ t \mapsto x^t$ is a strictly convex function, by Karamata's inequality, we have \begin{align*} x^k + x^{k+4} + x^{3k+2} + x^{3k+6} &< x^0 + x^2 + x^{4k+4} + x^{4k+6} \\ x^k(1+x^4)(1+x^{2k+2}) &< (1+x^2)(1+x^{4k+4}) \\ \frac{1+x^{2k+2}}{1+x^{4k+4}} &< \frac{1+x^2}{1+x^4} \cdot \frac{1}{x^k} \end{align*}for any nonnegative integer $ k$, since the sequence $ (4k+6,4k+4,2,0)$ strictly majorizes the sequence $ (3k+6,3k+2,k+4,k)$. (One can also prove this inequality by repeated application of weighted AM-GM.) Thus \begin{align*} \sum_{k=1}^n \frac{1+x^{2k}}{1+x^{4k}} &< \sum_{k=0}^{n-1} \frac{1+x^2}{1+x^4} \cdot \frac{1}{x^k} \\ &= \frac{1+x^2}{1+x^4} \cdot \frac{1/x^n - 1}{1/x-1} \\ &= \frac{(1+x^2)x(1-x^n)}{(1+x^4)x^n(1-x)} . \end{align*}Again by Karamata, $ (1+x^2)x = x+x^3 \le 1+x^4$, so \[ \frac{(1+x^2)x(1-x^n)}{(1+x^4)x^n(1-x)} \le \frac{(1-x^n)}{x^n(1-x)} = \frac{y^n}{x^n(1-x)} , \]since $ x^n + y^n = 1$. It now follows that \[ \biggl( \sum_{k=1}^n \frac{1+x^{2k}}{1+x^{4k}} \biggr) \cdot \biggl( \sum_{k=1}^n \frac{1+y^{2k}}{1+y^{4k}} \biggr) < \frac{y^n}{x^n(1-x)} \cdot \frac{x^n}{y^n(1-y)} = \frac{1}{(1-x)(1-y)}, \]as desired, since all quanties in question are positive. $ \blacksquare$ Nice

Let $0< n< 1.$ \[n< 1\implies n-n^4<1-n^3\iff \frac{1+n^2}{1+n^4}<\frac{1}{n}\]Thus, \[ \left(\sum^n_{k = 1} \frac {1 + x^{2k}}{1 + x^{4k}} \right) \cdot \left( \sum^n_{k = 1} \frac {1 + y^{2k}}{1 + y^{4k}} \right) < \left(\sum^n_{k = 1} \frac {1}{x^{k}} \right) \cdot \left( \sum^n_{k = 1} \frac {1}{y^{k}} \right)=\frac{1-x^n}{x^n(1-x)}\cdot \frac{1-y^n}{y^n(1-y)}=\frac{1}{1-x}\cdot \frac{1}{1-y}\]as desired.

awesomeming327. wrote: Let $0\le n< 1.$ \[n\le 1\implies n-n^4<1-n^3\iff \frac{1+n^2}{1+n^4}<\frac{1}{n}\]Thus, \[ \left(\sum^n_{k = 1} \frac {1 + x^{2k}}{1 + x^{4k}} \right) \cdot \left( \sum^n_{k = 1} \frac {1 + y^{2k}}{1 + y^{4k}} \right) \le \left(\sum^n_{k = 1} \frac {1}{x^{k}} \right) \cdot \left( \sum^n_{k = 1} \frac {1}{y^{k}} \right)=\frac{1-x^n}{x^n(1-x)}\cdot \frac{1-y^n}{y^n(1-y)}=\frac{1}{1-x}\cdot \frac{1}{1-y}\]as desired. The inequality is strict. $n=1$ doesn't work here. Maybe it's a typo.

Oops again fixed

$\begin{aligned}\left(\sum\limits_{k=1}^n\dfrac{1+x^{2k}}{1+x^{4k}}\right)\left(\sum\limits_{k=1}^n\dfrac{1+y^{2k}}{1+y^{4k}}\right) &=\left(\sum\limits_{k=1}^n\dfrac{x^k}{x^{2k}}\cdot\dfrac{x^k+\dfrac{1}{x^k}}{x^{2k}+\dfrac{1}{x^{2k}}}\right)\left(\sum\limits_{k=1}^n\dfrac{y^k}{y^{2k}}\cdot\dfrac{y^k+\dfrac{1}{y^k}}{y^{2k}+\dfrac{1}{y^{2k}}}\right)\\ &<\sum\limits_{k=1}^nx^{-k}\sum\limits_{k=1}^ny^{-k}=\dfrac{\sum\limits_{k=0}^{n-1}x^k}{x^n}\cdot \dfrac{\sum\limits_{k=0}^{n-1}y^k}{y^n}=\dfrac{1-x^n}{x^n(1-x)}\cdot\dfrac{1-y^n}{y^n(1-y)}<\frac{1}{(1-x)(1-y)}\end{aligned}$

The problem dies to the estimate $$\frac{1+x^{2k}}{1+x^{4k}} \leq x^{-k} \iff 1+x^{4k} \geq x^k + x^{3k}.$$Indeed this means that $$\left(\sum \frac{1+x^{2k}}{1+x^{4k}}\right)\left(\sum \frac{1+y^{2k}}{1+y^{4k}}\right) \leq x^{-1} \cdot \frac{1-x^{-n}}{1-x^{-1}} \cdot y^{-1} \cdot \frac{1-y^{-n}}{1-y^{-1}} = \frac 1{(1-x)(1-y)}.$$Note that equality cannot hold everywhere.

Claim: \[ \frac{1 + x^{2k}}{1 + x^{4k}} < x^{-k} \]Proof. Expands as \[ 1 + x^{2k} < x^{-k} + x^{3k} \]which follows by AM-GM. $\blacksquare$ As such, \[ \left(\sum_{k=1}^n \frac {1}{x^k}\right) \left(\sum_{k=1}^n \frac {1}{y^k} \right) < \frac{\frac{1}{x^n} - 1}{1 - x} \cdot \frac{\frac{1}{y^n} - 1}{1 - y} = \frac{1 - \frac{1 - x^n - y^n}{x^ny^n}}{(1-x)(1-y)} = \frac{1}{(1-x)(1-y)}. \]

Our claim is that \[\frac{1+t^2}{1+t^4} < \frac 1t\]for all $t \in (0,1)$, which can be proven by expanding to $(t-1)^2(t^2+t+1) > 0$. Substituting $x^k$ and $y^k$, we get \begin{align*} \left(\sum \frac{1+x^{2k}}{1+x^{4k}}\right)\left(\sum \frac{1+y^{2k}}{1+y^{4k}}\right) &< \left(\sum \frac {1}{x^k}\right) \left(\sum \frac {1}{y^k} \right) \\ &= \frac{1-x^n}{x^n(1-x)} \cdot \frac{1-y^n}{y^n(1-y)} = \frac{1}{(1-x)(1-y)}.~\blacksquare \end{align*}

Note that for $x<1$ we have $\frac{1+x^2}{1+x^4}<\frac 1x$. Thus, $$\left(\sum^n_{k = 1} \frac {1 + x^{2k}}{1 + x^{4k}} \right)\left( \sum^n_{k = 1} \frac {1 + y^{2k}}{1 + y^{4k}} \right)<\left(\sum^n_{k = 1} \frac {1}{x^{k}} \right)\left( \sum^n_{k = 1} \frac {1}{y^{k}} \right)=\frac{(1-x^n)(1-y^n)}{x^ny^n(1-x)(1-y)}=\frac{1}{(1-x)(1-y)}. $$

Note that \[\sum_{k=1}^n\frac{1+x^{2k}}{1+x^{4k}}\le\sum_{k=1}^n\frac{1}{x^k}=\frac{\frac{1}{x^{n}}-1}{1-x}.\]The equality case is when $x^k=1$, which is impossible, so the inequality is strict. Then indeed \[\frac{\frac{1}{x^n}-1}{1-x}\cdot\frac{\frac{1}{y^n}-1}{1-y}=\frac{1}{(1-x)(1-y)}\]because $\frac{1}{x^n y^n}=\frac{1}{x^n}+\frac{1}{y^n}$. $\blacksquare$