Shouldnt the sigma condition be $1 \le i < j \le n$.

GorgonMathDota wrote: Shouldnt the sigma condition be $1 \le i < j \le n$. You are right. sqing wrote: Let $a_1,a_2,\cdots,a_n (n\ge 2)$ be positive numbers such that $a_1\leq a_2 \leq \cdots \leq a_n .$ Prove that $$\sum_{1\leq i\leq j \leq n} (a_i+a_j)^2\left(\frac{1}{i^2}+\frac{1}{j^2}\right)\geq 4(n-1)\sum_{i=1}^{n}\frac{a^2_i}{i^2}.$$ Please correct it.

If $n=2$, it is $(a_2-a_1)(a_2+11a_1)\geq 0$. If $n>k\geq 1$ and $a_{k+1}=a_{k+2}=\cdots=a_n=x$, The LHS becomes $$\sum_{j=k+1}^{n}\sum_{i=1}^{k}(x_j+a_i)^2\cdot\frac{1}{i^2} +\sum_{i=1}^{k}\sum_{j=k+1}^{n}(x_j+a_i)^2\cdot\frac{1}{j^2}+$$$$+\sum_{k+1\leq j<l\leq n}(x_j+x_l)^2(\frac{1}{j^2}+\frac{1}{l^2})+\text{nothing with }x$$then we take derivative of LHS-RHS in $x$. The derivative is larger than $$2x(n-k)\sum_{i=1}^{k}\frac{1}{i^2}+2xk\sum_{j=k+1}^{n}\frac{1}{j^2}+$$$$+8x(n-k-1)\sum_{j=k+1}^{n}\frac{1}{j^2}-8x(n-1)\sum_{i=k+1}^{n}\frac{1}{j^2}$$The coefficient of $x$ is $$2(n-k)\sum_{i=1}^{k}\frac{1}{i^2}-6k\sum_{j=k+1}^{n}\frac{1}{j^2}$$when we know $\sum_{i=1}^{k}\frac{1}{i^2}\geq \frac{3}{2}-\frac{1}{k+1}$ and $\sum_{j=k+1}^{n}\frac{1}{j^2}\leq (\frac{1}{k}-\frac{1}{n})$ Then this coefficient is larger than or equal to $$2(n-k)\cdot(\frac{3}{2}-\frac{1}{k+1})-6k(\frac{1}{k}-\frac{1}{n})$$ which is $$\frac{(n-k)(3kn-6k+n-6)}{(k+1)n}$$ But when $n\geq 3$, then $3kn-6k+n-6\geq 3k-3\geq 0$ because $k\geq 1$. Hence the derivative in $x$ is $\geq 0$ when $n\geq 3$. Then take $k=n-1, n-2,\cdots, 1$, it tells us \begin{align*} &f(a_1,a_2,\cdots,a_{n-1},a_n)\\ \geq&f(a_1,a_2,\cdots,a_{n-1},a_{n-1})\\ \geq &f(a_1,a_2,\cdots,a_{n-2},a_{n-2},a_{n-2})\\ &\cdots\\ \geq &f(a_1,a_1,\cdots,a_1)\\ =&0 \end{align*}where $f(a_1,\cdots,a_n)=LHS-RHS$.

yleo wrote: If $n=2$, it is $(a_2-a_1)(a_2+11a_1)\geq 0$. If $n>k\geq 1$ and $a_{k+1}=a_{k+2}=\cdots=a_n=x$, The LHS becomes $$\sum_{j=k+1}^{n}\sum_{i=1}^{k}(x_j+a_i)^2\cdot\frac{1}{i^2} +\sum_{i=1}^{k}\sum_{j=k+1}^{n}(x_j+a_i)^2\cdot\frac{1}{j^2}+$$$$+\sum_{k+1\leq j<l\leq n}(x_j+x_l)^2(\frac{1}{j^2}+\frac{1}{l^2})+\text{nothing with }x$$then we take derivative of LHS-RHS in $x$. The derivative is larger than $$2x(n-k)\sum_{i=1}^{k}\frac{1}{i^2}+2xk\sum_{j=k+1}^{n}\frac{1}{j^2}+$$$$+8x(n-k-1)\sum_{j=k+1}^{n}\frac{1}{j^2}-8x(n-1)\sum_{i=k+1}^{n}\frac{1}{j^2}$$The coefficient of $x$ is $$2(n-k)\sum_{i=1}^{k}\frac{1}{i^2}-6k\sum_{j=k+1}^{n}\frac{1}{j^2}$$when we know $\sum_{i=1}^{k}\frac{1}{i^2}\geq \frac{3}{2}-\frac{1}{k+1}$ and $\sum_{j=k+1}^{n}\frac{1}{j^2}\leq (\frac{1}{k}-\frac{1}{n})$ Then this coefficient is larger than or equal to $$2(n-k)\cdot(\frac{3}{2}-\frac{1}{k+1})-6k(\frac{1}{k}-\frac{1}{n})$$ which is $$\frac{(n-k)(3kn-6k+n-6)}{(k+1)n}$$ But when $n\geq 3$, then $3kn-6k+n-6\geq 3k-3\geq 0$ because $k\geq 1$. Hence the derivative in $x$ is $\geq 0$ when $n\geq 3$. Then take $k=n-1, n-2,\cdots, 1$, it tells us \begin{align*} &f(a_1,a_2,\cdots,a_{n-1},a_n)\\ \geq&f(a_1,a_2,\cdots,a_{n-1},a_{n-1})\\ \geq &f(a_1,a_2,\cdots,a_{n-2},a_{n-2},a_{n-2})\\ &\cdots\\ \geq &f(a_1,a_1,\cdots,a_1)\\ =&0 \end{align*}where $f(a_1,\cdots,a_n)=LHS-RHS$. Where can I learn these sigma equations?

I can't translate because my English is taught by my physical education teacher(我真的不会翻译了，我英语是体育老师教的): 我们先做恒等变形, 只要证明: $$\sum_{1\leq i\neq j \leq n} (a_i+a_j)^2\left(\frac{1}{i^2}+\frac{1}{j^2}\right)\geq 8(n-1)\sum_{i=1}^{n}\frac{a^2_i}{i^2},$$即: $$\sum_{i=1}^n\sum_{j=1}^n (a_i+a_j)^2\left(\frac{1}{i^2}+\frac{1}{j^2}\right)\geq 8n\sum_{i=1}^{n}\frac{a^2_i}{i^2},$$即: $$\sum_{i=1}^n\sum_{j=1}^n (a_i+a_j)^2\frac{1}{i^2}\geq 4n\sum_{i=1}^{n}\frac{a^2_i}{i^2},$$即: $$ f =2\sum_{j=1}^n a_j \sum_{i=1}^n \frac{a_i}{i^2}+\sum_{i=1}^n \frac{1}{i^2} \sum_{j=1}^n a_j^2-3n\sum_{i=1}^{n}\frac{a^2_i}{i^2} =\sum_{i=1}^n [(\sum_{k=1}^n\frac{1}{k^2})-\frac{3n-2}{i^2}]a_i^2+\frac{2[(\sum_{k=1}^n a_k)-a_i]}{i^2}a_i \geq 0.$$ 注意到: 当$(\sum_{k=1}^n\frac{1}{k^2})-\frac{3n-2}{i^2} \leq0$时,$f$关于$a_i$上凸(开口向下); 当$(\sum_{k=1}^n\frac{1}{k^2})-\frac{3n-2}{i^2} >0$时,$f$关于$a_i$下凸(开口向上), 而此时对称轴(极值点)为负, 我们设$f$关于$a_1, a_2, \cdots, a_l$是上凸函数, 关于$a_{l+1}, \cdots, a_n$是下凸函数, $f$取最小值时必满足: $a_{1}=a_2= \cdots= a_k$, $a_{k+1}= \cdots= a_l$, $a_{l+1}= \cdots= a_n$, 再注意到$0<a_l \leq a_{l+1}$, 且对称轴(极值点)为负, 这样一来, $f$在$a_l=a_{l+1}= \cdots= a_n$时取最小值(把$a_{l+1}$调整到$a_l$, $f$更小且不可再调). 于是我们得到, $f$ 在$a_1=a_2=\cdots=a_k=x \leq a_{k+1}=a_{k+2}=\cdots=a_n=y$ 时取到最小值. 我们记$\sum\limits_{i=1}^k \frac{1}{i^2}=A$, $B=\sum\limits_{i=k+1}^n \frac{1}{i^2}<\frac{1}{k(k+1)}+\cdots+\frac{1}{(n-1)n}<\frac{1}{k}$, 此时, $$ f =[kB-3A(n-k)]x^2+[A(n-k)-3kB]y^2+2[kB+(n-k)A]xy =(ux-vy)(x-y). $$其中, $u=kB-3A(n-k)$, $v=A(n-k)-3kB$. 由于$x \leq y$, $ux-vy \leq (u-v)y$, 下面我们只要说明$u-y \leq 0$: 当$k=n$时结论显然成立; 当$k \leq n-1$时, $u-v=4(kB-An+Ak) \leq 4(kA-nA+1) \leq 4(1-A) \leq 0$. 我们粗暴地完成了证明.

WypHxr wrote: I can't translate because my English is taught by my physical education teacher(我真的不会翻译了，我英语是体育老师教的): 我们先做恒等变形, 只要证明: $$\sum_{1\leq i\neq j \leq n} (a_i+a_j)^2\left(\frac{1}{i^2}+\frac{1}{j^2}\right)\geq 8(n-1)\sum_{i=1}^{n}\frac{a^2_i}{i^2},$$即: $$\sum_{i=1}^n\sum_{j=1}^n (a_i+a_j)^2\left(\frac{1}{i^2}+\frac{1}{j^2}\right)\geq 8n\sum_{i=1}^{n}\frac{a^2_i}{i^2},$$即: $$\sum_{i=1}^n\sum_{j=1}^n (a_i+a_j)^2\frac{1}{i^2}\geq 4n\sum_{i=1}^{n}\frac{a^2_i}{i^2},$$即: $$ f =2\sum_{j=1}^n a_j \sum_{i=1}^n \frac{a_i}{i^2}+\sum_{i=1}^n \frac{1}{i^2} \sum_{j=1}^n a_j^2-3n\sum_{i=1}^{n}\frac{a^2_i}{i^2} =\sum_{i=1}^n [(\sum_{k=1}^n\frac{1}{k^2})-\frac{3n-2}{i^2}]a_i^2+\frac{2[(\sum_{k=1}^n a_k)-a_i]}{i^2}a_i \geq 0.$$ 注意到: 当$(\sum_{k=1}^n\frac{1}{k^2})-\frac{3n-2}{i^2} \leq0$时,$f$关于$a_i$上凸(开口向下); 当$(\sum_{k=1}^n\frac{1}{k^2})-\frac{3n-2}{i^2} >0$时,$f$关于$a_i$下凸(开口向上), 而此时对称轴(极值点)为负, 我们设$f$关于$a_1, a_2, \cdots, a_l$是上凸函数, 关于$a_{l+1}, \cdots, a_n$是下凸函数, $f$取最小值时必满足: $a_{1}=a_2= \cdots= a_k$, $a_{k+1}= \cdots= a_l$, $a_{l+1}= \cdots= a_n$, 再注意到$0<a_l \leq a_{l+1}$, 且对称轴(极值点)为负, 这样一来, $f$在$a_l=a_{l+1}= \cdots= a_n$时取最小值(把$a_{l+1}$调整到$a_l$, $f$更小且不可再调). 于是我们得到, $f$ 在$a_1=a_2=\cdots=a_k=x \leq a_{k+1}=a_{k+2}=\cdots=a_n=y$ 时取到最小值. 我们记$\sum\limits_{i=1}^k \frac{1}{i^2}=A$, $B=\sum\limits_{i=k+1}^n \frac{1}{i^2}<\frac{1}{k(k+1)}+\cdots+\frac{1}{(n-1)n}<\frac{1}{k}$, 此时, $$ f =[kB-3A(n-k)]x^2+[A(n-k)-3kB]y^2+2[kB+(n-k)A]xy =(ux-vy)(x-y). $$其中, $u=kB-3A(n-k)$, $v=A(n-k)-3kB$. 由于$x \leq y$, $ux-vy \leq (u-v)y$, 下面我们只要说明$u-y \leq 0$: 当$k=n$时结论显然成立; 当$k \leq n-1$时, $u-v=4(kB-An+Ak) \leq 4(kA-nA+1) \leq 4(1-A) \leq 0$. 我们粗暴地完成了证明. 为什么对于上凸的部分可以调到仅有两个值？ 我只能做到$a_i \in \{a_{i-1},a_{i+1}\}$