48th International Mathematical Olympiad Vietnam 2007 Shortlisted Problems with Solutions

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 $t=y^k$, $$0<\sum _{k=1}^n\frac{1+x^{2k}}{1+x^{4k}}<\sum _{k=1}^n\frac{1}{x^k}=\frac{1-x^n}{x^n(1-x)}\text{ and }0<\sum _{k=1}^n\frac{1+y^{2k}}{1+y^{4k}}<\sum _{k=1}^n\frac{1}{y^k}=\frac{1-y^n}{y^n(1-y)}.$$ Since $1-y^n=x^n$ and $1-x^n=y^n$, $$\frac{1-x^n}{x^n(1-x)}=\frac{y^n}{x^n(1-x)},\frac{1-y^n}{y^n(1-y)}=\frac{x^n}{y^n(1-y)}$$ and therefore $$\left(\sum _{k=1}^n\frac{1+x^{2k}}{1+x^{4k}}\right)\left(\sum _{k=1}^n\frac{1+y^{2k}}{1+y^{4k}}\right)<\frac{y^n}{x^n(1-x)}\cdot \frac{x^n}{y^n(1-y)}=\frac{1}{(1-x)(1-y)}.$$

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Alternative solution.

We prove $$\begin{array}{c}\left(\sum _{k=1}^n\frac{1+x^{2k}}{1+x^{4k}}\right)\left(\sum _{k=1}^n\frac{1+y^{2k}}{1+y^{4k}}\right)<\frac{{\left(\frac{1+\sqrt{2}}{2}\ln 2\right)}^2}{(1-x)(1-y)}<\frac{0.7001}{(1-x)(1-y)}\end{array}\text{\hspace{1em}(1)}$$ The idea is to estimate each term on the left-hand side with the same constant. To find the upper bound for the expression $\frac{1+x^{2k}}{1+x^{4k}}$, consider the function $f(t)=\frac{1+t}{1+t^2}$ in interval $(0,1)$. Since $$f^{\prime }(t)=\frac{1-2t-t^2}{{\left(1+t^2\right)}^2}=\frac{(\sqrt{2}+1+t)(\sqrt{2}-1-t)}{{\left(1+t^2\right)}^2}$$ 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 $$f(t)=\frac{1+t}{1+t^2}\le f\left(t_0\right)=\frac{1+\sqrt{2}}{2}=\alpha$$ Applying this to each term on the left-hand side of (1), we obtain $$\begin{array}{c}\left(\sum _{k=1}^n\frac{1+x^{2k}}{1+x^{4k}}\right)\left(\sum _{k=1}^n\frac{1+y^{2k}}{1+y^{4k}}\right)\le n\alpha \cdot n\alpha =(n\alpha {)}^2.\end{array}\text{\hspace{1em}(2)}$$ 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 $$-n{g}^{\prime }(t)=\frac{t^{1/n-1}}{1-t^{1/n}}-\frac{s^{1/n-1}}{1-s^{1/n}}=\frac{1}{st}\left(\frac{(1-t)t^{1/n}}{1-t^{1/n}}-\frac{(1-s)s^{1/n}}{1-s^{1/n}}\right)=\frac{h(t)-h(s)}{st}.$$ 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 $$g^{\prime }(t)>0⟺h(t)<h(s)⟺t<s=1-t⟺t<\frac{1}{2}.$$ Then, the maximum of $g(t)$ in ( 0,1 ) is attained at point $t_1=1/2$ and therefore $$g(t)\le g\left(\frac{1}{2}\right)=2\ln \left(1-2^{-1/n}\right),t\in (0,1)$$ Substituting $t=x^n$, we have $1-t=y^n,(1-x)(1-y)=\exp g(t)$ and hence $$\begin{array}{c}(1-x)(1-y)=\exp g(t)\le {\left(1-2^{-1/n}\right)}^2\end{array}\text{\hspace{1em}(3)}$$ Combining (2) and (3), we get $$\left(\sum _{k=1}^n\frac{1+x^{2k}}{1+x^{4k}}\right)\left(\sum _{k=1}^n\frac{1+y^{2k}}{1+y^{4k}}\right)\le (\alpha n{)}^2\cdot 1\le (\alpha n{)}^2\frac{{\left(1-2^{-1/n}\right)}^2}{(1-x)(1-y)}=\frac{{\left(\alpha n\left(1-2^{-1/n}\right)\right)}^2}{(1-x)(1-y)}.$$ 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, $$\left(\sum _{k=1}^n\frac{1+x^{2k}}{1+x^{4k}}\right)\left(\sum _{k=1}^n\frac{1+y^{2k}}{1+y^{4k}}\right)<\frac{{\left(\frac{1+\sqrt{2}}{2}\ln 2\right)}^2}{(1-x)(1-y)}$$