Let $a_1\geq a_2\geq \ldots \geq a_n$ and $b_1\geq b_2 \geq \ldots \geq b_n$ be two sequences with $\sum_{i=1}^n a_i=0$ and $\sum_{i=1}^n b_i=0.$ I want to prove if there exists $1\leq j\leq n$ such that $(a_j-b_j)^2>m$, then $$\sum_{i=1}^n (a_i-b_i)^2>2m.$$
I assumed that $\sum_{i=1}^n (a_i-b_i)^2\leq 2m$ and tried to use their property to get the result but it didn't work. Any kind of suggestion is appreciated. Is there any books about inequalities like these? Thanks to everyone for the help.
Let us assume that $n\geq 3$ and that $j$ is neither $1$ or $n$. By the convexity of $x\mapsto x^2$, $$ \sum_{k=1}^{n}(a_k-b_k)^2 \geq \sum_{k=1}^{j-1}(a_k'-b_k')^2+(a_j-b_j)^2+\sum_{k=j+1}^{n}(a_k''-b_k'')^2 $$ where $a_k'$ is the average of $a_1,\ldots,a_{j-1}$, $b_k'$ is the average of $b_1,\ldots,b_{j-1}$, $a_k''$ is the average of $a_{j+1},\ldots,a_{n}$ and $b_k''$ is the average of $b_{j+1},\ldots,b_n$. This averaging process preserves the monotonicity and the constraints on the sum of our sequences, hence we may assume without loss of generality that our sequences are constant for $k<j$ and $k>j$, attaining the values $A_1=a_k',A_2=a_j,A_3=a_k''$ and $B_1=b_k',B_2=b_j,B_3=b_k''$. We want to prove that
$$ (j-1)(A_1-B_1)^2 + (n-j)(A_3-B_3)^2 $$ is $\geq m$, knowing that $A_1\geq A_2\geq A_3$, $B_1\geq B_2\geq B_3$, $|A_2-B_2|>\sqrt{m}$ and $$ (j-1)A_1+A_2+(n-j)A_3 = (j-1)B_1+B_2+(n-j)B_3 = 0.$$ Up to exchanging $A$ and $B$ we may also assume that $A_2>B_2$. Let $B_2=x$, $A_2=x+\tau$ (with $\tau>\sqrt{m}$), $B_1=x+\mu$ (with $\mu\geq 0$), $A_1=x+\tau+\lambda$ with $\lambda\geq 0$. We have $$ A_3 = -\frac{(j-1)A_1+A_2}{n-j}= -\frac{(j-1)(x+\tau+\lambda)+(x+\tau)}{n-j}\leq A_2=(x+\tau), $$ so $$ -(j-1)\lambda\leq n(x+\tau)$$ and similarly $$ B_3 = -\frac{(j-1)B_1+B_2}{n-j}= -\frac{(j-1)(x+\mu)+x}{n-j}\leq B_2=x, $$ so $$-(j-1)\mu\leq nx.$$ Now $$ (n-j)(A_3-B_3)=-(j-1)(A_1-B_1)-(A_2-B_2) $$ leads to $$ (n-j)(A_3-B_3)^2 = \frac{1}{n-j}\left[(j-1)(\tau+\lambda-\mu)+\tau\right]^2 $$ and since $$ (j-1)(A_1-B_1)^2 = (j-1)(\tau+\lambda-\mu)^2$$ we just have to show that the previous constraints ensure $$ (j-1)(\tau+\lambda-\mu)^2+\frac{1}{n-j}\left[(j-1)(\tau+\lambda-\mu)+\tau\right]^2\geq \tau^2. $$ Not an elegant way, but it should do the job. The initial assumption $1<j<n$ is not really restrictive since we may always insert $a_0=b_0=M$ and $a_{n+1}=b_{n+1}=-M$ into valid sequences, getting valid sequences with length increased by $2$.