China Girls' Mathematical Olympiad

Comment: Expanding the left-hand side of the desired inequality gives $$\begin{array}{rl}& \sum _{k=1}^n{\left(1-\frac{k}{{\sum }_{i=1}^{n}i{x}_i^2}\right)}^2\cdot \frac{x_k^2}{k}\\ & =\sum _{k=1}^n\frac{x_k^2}{k}-\sum _{k=1}^n\frac{2x_k^2}{{\sum }_{i=1}^{n}i{x}_i^2}+\sum _{k=1}^n\frac{k{x}_k^2}{{\left({\sum }_{i=1}^{n}i{x}_i^2\right)}^2}\\ & =\sum _{k=1}^n\frac{x_k^2}{k}-\frac{1}{{\sum }_{i=1}^{n}i{x}_i^2}\sum _{k=1}^n\frac{1}{2x_k^2}+\frac{1}{{\left({\sum }_{i=1}^{n}i{x}_i^2\right)}^2}\sum _{k=1}^{n}k{x}_k^2\\ & =\sum _{k=1}^n\frac{x_k^2}{k}-\frac{2}{{\sum }_{i=1}^{n}i{x}_i^2}+\frac{1}{{\sum }_{i=1}^{n}i{x}_i^2}\\ & =\sum _{k=1}^n\frac{x_k^2}{k}-\frac{1}{{\sum }_{i=1}^{n}i{x}_i^2}.\end{array}$$ We want to show that $$\sum _{k=1}^n\frac{x_k^2}{k}-\frac{1}{{\sum }_{i=1}^{n}i{x}_i^2}\le {\left(\frac{n-1}{n+2}\right)}^2\sum _{k=1}^n\frac{x_k^2}{k}=\sum _{k=1}^n\frac{x_k^2}{k}-\frac{4n}{(n+1{)}^2}\sum _{k=1}^n\frac{x_k^2}{k}$$ or $$\frac{4n}{(n+1{)}^2}\sum _{k=1}^n\frac{x_k^2}{k}\le \frac{1}{{\sum }_{i=1}^{n}i{x}_i^2},$$ that is, $$\left(\sum _{k=1}^n\frac{x_k^2}{k}\right)\left(\sum _{k=1}^{n}k{x}_k^2\right)\le \frac{(n+1{)}^2}{4n}.\stackrel{◯}{1}$$ We present two proofs of the above inequality.

Solution 1. We rewrite 1 as $$4n\left(\sum _{k=1}^n\frac{x_k^2}{k}\right)\left(\sum _{k=1}^{n}k{x}_k^2\right)\le (n+1{)}^2.$$ By the AM-GM inequality, we have $$\begin{array}{rl}4n\left(\sum _{k=1}^n\frac{x_k^2}{k}\right)\left(\sum _{k=1}^{n}k{x}_k^2\right)& =4\left(\sum _{k=1}^n\frac{n{x}_k^2}{k}\right)\left(\sum _{k=1}^{n}k{x}_k^2\right)\\ & \le {\left(\sum _{k=1}^n\frac{n{x}_k^2}{k}+\sum _{k=1}^{n}k{x}_k^2\right)}^2\\ & ={\left(\sum _{k=1}^n\left(\frac{n}{k}+k\right)x_k^2\right)}^2.\end{array}$$ It suffices to show that $$\frac{n}{k}+k\le n+1$$ or $$0\le nk+k-k^2-n=(n-k)(k-1),$$ which is evident. Now, we consider the equality case. Note that the last inequality is strict for $1<k<n$. Hence, we must have $x_2=\cdots =x_{n-1}=0$. For the AM-GM inequality to hold, we must have $$\sum _{k=1}^n\frac{n{x}_k^2}{k}=\sum _{k=1}^{n}k{x}_k^2\text{ or }n{x}_1^2+x_n^2=x_1^2+n{x}_n^2$$ with $x_1^2+x_n^2=1$. We must have $x_1^2=x_n^2=\frac{1}{2}$ and $x_2=\cdots =x_{n-1}=0$.

Solution 2. (By Lynnelle Ye) We write ④ as $$(n+1{)}^2-4n\left(\sum _{k=1}^n\frac{x_k^2}{k}\right)\left(\sum _{k=1}^{n}k{x}_k^2\right)\ge 0.$$ Note that the left-hand side of the above inequality is the discriminant of the quadratic (in $t$): $$f(t)=n(x_1^2+2x_2^2+\cdots +n{x}_n^2)t^2-(n+1)t+\left(x_1^2+\frac{x_2^2}{2}+\cdots +\frac{x_n^2}{n}\right).$$ It suffices to show that $f(t)$ has a real root. Because the leading coefficient of $f(x)$ is $n(x_1^2+2x_2^2+\cdots +n{x}_n^2)$, which is positive, it remains to be shown that $f(t)\le 0$ for some $t$. Because $x_1^2+\cdots +x_n^2=1$, we have $$\begin{array}{rl}f(t)& =n(x_1^2+2x_2^2+\cdots +n{x}_n^2)t^2-(n+1)(x_1^2+x_2^2+\cdots +x_n^2)t+\left(x_1^2+\frac{x_2^2}{2}+\cdots +\frac{x_n^2}{n}\right)t\\ & =\sum _{k=1}^n\left(nk{x}_k^2{t}^2-(n+1)x_k^{2}t+\frac{x_k^2}{k}\right)\\ & =\sum _{k=1}^n{x}_k^2(nkt-1)\left(t-\frac{1}{k}\right).\end{array}$$ It is easy to see that $f(\frac{1}{n})\le 0$ because for $t=\frac{1}{n}$, each summand $$x_k^2(nkt-1)\left(t-\frac{1}{k}\right)=\frac{x_k^2(k-1)(k-n)}{nk}\le 0,$$ completing our proof. For the equality case of the given inequality, we must have equality case for the above inequality for every $k$; that is, $x_k=0$ for $2\le k\le n-1$. It is then not difficult to obtain that $$ x_1^2 = x_2^2 = \frac{1}{2}. \quad \square