# Find optimal $k$ such that $(\sum_{i=1}^n a_i(x_i^2-y_i^2))^2\le k^2\sum_{i=1}^n(x_i-y_i)^2$, with $\sum_{i=1}^n x_i^2, \sum_{i=1}^n y_i^2\le1$

Let $$\{a_i\}$$ be $$n$$ real numbers. It can be shown that there is $$k\ge 0$$ such that for all $$x_i,y_i\in\mathbb{R}$$ satisfying $$\sum_{i=1}^n x_i^2\le 1$$ and $$\sum_{i=1}^n y_i^2\le 1$$, $$(\sum_{i=1}^n a_i(x_i^2-y_i^2))^2\le k^2 \sum_{i=1}^n (x_i-y_i)^2.$$ I proved the above with $$k=2\sqrt{\sum_{i=1}^n {a_i^2}}$$, but I suspect that $$k=2\max\{|a_i|\}$$ is the optimal value for $$k$$. I struggle to produce a proof.

• Try first for $n=1.$ Then extend to any $n,$ using CS. And you are not asked for "an optimal value". Dec 3, 2022 at 6:16
• @AnneBauval I actually managed to prove the existence of $k$, but now I would like to find the optimal $k$. Changed the question a bit. Dec 3, 2022 at 6:29
• Again: instead (or before) proving $k=2\max\{a_i\}$ is "an optimal value", prove it is a value. Try when all $a_i$'s are equal. Dec 3, 2022 at 6:39
• @AnneBauval It doesn't seem that the equality is always attainable for $k=2\max\{|a_i|\}$. I'm considering the quotient $(\sum a(x^2-y^2))^2/(\sum (x-y)^2)$, but I don't know how to maximize this quotient. Dec 3, 2022 at 7:09
• by using C-S, $k^2\geq\sum a_i^2(x_i+y_i)^2$ is sufficient for above inequality to hold. And from it you can show $k^2\geq 4\max a_i^2$ would be sufficient. It remains to check the choice $x_i=1, y_i=1-\epsilon$ where $i$ is the index for maximal $a_i$ to justify your optimal argument. Dec 3, 2022 at 7:51

If all $$a_i$$ are $$0$$, then clearly $$k=0$$ is the optimal value. Otherwise, there is $$a_j$$ such that $$|a_j|=\max\{|a_i|\}>0$$. Choose $$k:=2|a_j|$$. Then \begin{align*} \left( \sum_{i=1}^{n} a_i(x_i^2-y_i^2) \right)^2 &= \left( \sum_{i=1}^{n} a_i(x_i+y_i)(x_i-y_i) \right)^2 \\ &\le \sum_{i=1}^{n} a_i^2(x_i+y_i)^2 \sum_{i=1}^{n} (x_i-y_i)^2 & \text{Cauchy's inequality} \\ &= a_j^2 \sum_{i=1}^{n} \frac{a_i^2}{a_j^2} (x_i+y_i)^2 \sum_{i=1}^{n} (x_i-y_i)^2 & a_j\neq 0\\ &\le a_j^2 \sum_{i=1}^{n} (x_i+y_i)^2 \sum_{i=1}^{n} (x_i-y_i)^2 & \forall i:a_j^2\ge a_i^2\\ &\le a_j^2 \left( \sqrt{\sum_{i=1}^{n} x_i^2} + \sqrt{\sum_{i=1}^{n} y_i^2} \right)^2 \sum_{i=1}^{n} (x_i-y_i)^2 & \text{Triangle inequality} \\ &\le 4a_j^2 \sum_{i=1}^{n} (x_i-y_i)^2 & \sum_{i=1}^{n} x_i^2 \le 1,\sum_{i=1}^{n} y_i^2\le 1 \\ &= k^2 \sum_{i=1}^{n} (x_i-y_i)^2. \end{align*} We now prove that $$k$$ is optimal. Let $$0. Choose $$x_j:=1$$ and $$y_j:=1-r$$, and all other $$x_i,y_i:=0$$. Then \begin{align*} \lim_{r\to 0^+} \frac{(\sum_{i=1}^{n} a_i(x_i^2-y_i^2))^2}{\sum_{i=1}^{n} (x_i-y_i)^2} &= \lim_{r\to 0^+} \frac{a_j^2(1-(1-r)^2)^2}{r^2} \\ &= 4a_j^2 \\ &= k^2. \end{align*} This then shows that $$k^2=\sup\left\{\frac{(\sum_{i=1}^{n} a_i(x_i^2-y_i^2))^2}{\sum_{i=1}^{n} (x_i-y_i)^2}:x_i,y_i\in\mathbb{R};\sum_{i=1}^{n}x_i^2\le 1,\sum_{i=1}^{n}y_i^2\le 1 ;\sum_{i=1}^{n} (x_i-y_i)^2\neq 0\right\},$$ so $$k$$ is optimal.