IMO 2016 Shortlisted Problems

We first show that $a=\frac{4}{9}$ is admissible. For each $2⩽k⩽n$, by the Cauchy-Schwarz Inequality, we have $$\left(x_{k-1}+\left(x_k-x_{k-1}\right)\right)\left(\frac{(k-1{)}^2}{x_{k-1}}+\frac{3^2}{x_k-x_{k-1}}\right)⩾(k-1+3{)}^2$$ which can be rewritten as $$\frac{9}{x_k-x_{k-1}}⩾\frac{(k+2{)}^2}{x_k}-\frac{(k-1{)}^2}{x_{k-1}}\text{\hspace{1em}(2)}$$ Summing (2) over $k=2,3,\dots ,n$ and adding $\frac{9}{x_1}$ to both sides, we have $$9\sum _{k=1}^n\frac{1}{x_k-x_{k-1}}⩾4\sum _{k=1}^n\frac{k+1}{x_k}+\frac{n^2}{x_n}>4\sum _{k=1}^n\frac{k+1}{x_k}$$ This shows (1) holds for $a=\frac{4}{9}$.

Next, we show that $a=\frac{4}{9}$ is the optimal choice. Consider the sequence defined by $x_0=0$ and $x_k=x_{k-1}+k(k+1)$ for $k⩾1$, that is, $x_k=\frac{1}{3}k(k+1)(k+2)$. Then the left-hand side of (1) equals $$\sum _{k=1}^n\frac{1}{k(k+1)}=\sum _{k=1}^n\left(\frac{1}{k}-\frac{1}{k+1}\right)=1-\frac{1}{n+1}$$ while the right-hand side equals $$a\sum _{k=1}^n\frac{k+1}{x_k}=3a\sum _{k=1}^n\frac{1}{k(k+2)}=\frac{3}{2}a\sum _{k=1}^n\left(\frac{1}{k}-\frac{1}{k+2}\right)=\frac{3}{2}\left(1+\frac{1}{2}-\frac{1}{n+1}-\frac{1}{n+2}\right)a.$$ When $n$ tends to infinity, the left-hand side tends to 1 while the right-hand side tends to $\frac{9}{4}a$. Therefore $a$ has to be at most $\frac{4}{9}$.

Hence the largest value of $a$ is $\frac{4}{9}$.

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Alternative solution.

We shall give an alternative method to establish (1) with $a=\frac{4}{9}$. We define $y_k=x_k-x_{k-1}>0$ for $1⩽k⩽n$. By the Cauchy-Schwarz Inequality, for $1⩽k⩽n$, we have $$\left(y_1+y_2+\cdots +y_k\right)\left(\sum _{j=1}^k\frac{1}{y_j}{\left(\genfrac{}{}{0ex}{}{j+1}{2}\right)}^2\right)⩾{\left(\left(\genfrac{}{}{0ex}{}{2}{2}\right)+\left(\genfrac{}{}{0ex}{}{3}{2}\right)+\cdots +\left(\genfrac{}{}{0ex}{}{k+1}{2}\right)\right)}^2={\left(\genfrac{}{}{0ex}{}{k+2}{3}\right)}^2.$$ This can be rewritten as $$\frac{k+1}{y_1+y_2+\cdots +y_k}⩽\frac{36}{k^2(k+1)(k+2{)}^2}\left(\sum _{j=1}^k\frac{1}{y_j}{\left(\genfrac{}{}{0ex}{}{j+1}{2}\right)}^2\right)\text{\hspace{1em}(3)}$$ Summing (3) over $k=1,2,\dots ,n$, we get $$\frac{2}{y_1}+\frac{3}{y_1+y_2}+\cdots +\frac{n+1}{y_1+y_2+\cdots +y_n}⩽\frac{c_1}{y_1}+\frac{c_2}{y_2}+\cdots +\frac{c_n}{y_n}\text{\hspace{1em}(4)}$$ where for $1⩽m⩽n$, $$\begin{array}{rl}{c}_m& =36{\left(\genfrac{}{}{0ex}{}{m+1}{2}\right)}^2\sum _{k=m}^n\frac{1}{k^2(k+1)(k+2{)}^2}\\ & =\frac{9m^2(m+1{)}^2}{4}\sum _{k=m}^n\left(\frac{1}{k^2(k+1{)}^2}-\frac{1}{(k+1{)}^2(k+2{)}^2}\right)\\ & =\frac{9m^2(m+1{)}^2}{4}\left(\frac{1}{m^2(m+1{)}^2}-\frac{1}{(n+1{)}^2(n+2{)}^2}\right)<\frac{9}{4}\end{array}$$ From (4), the inequality (1) holds for $a=\frac{4}{9}$. This is also the upper bound as can be verified in the same way as Solution 1.