Proof of an equality with products

Question

Let $\lambda_{1},\lambda_{2},\cdots,\lambda_{n}$ be $n$ distinct real numbers. For $r=1,2,\cdots,n$, show that: $$L(r)=\sum_{p,q=1,p\neq q;p,q\neq r}\dfrac{1}{(\lambda_{r}-\lambda_{p})(\lambda_{r}-\lambda_{q})\cdot\prod_{k=1,k\neq p}^{n}(\lambda_{k}-\lambda_{p})\prod_{l=1,l\neq q}(\lambda_{l}-\lambda_{q})}<0$$ I was told that the inequality comes from the result of an article (but I was not told which article). It looks very difficult to prove to me, but there might be an elementary solution.

Answer 1

it is not a complete proof, edit for error correction, but too long to write in comment, i think you can find something in this observation

$$L(r)=\sum_{p,q=1,p\neq q;p,q\neq r}^{p,q=n}\dfrac{1}{(\lambda_{r}-\lambda_{p})(\lambda_{r}-\lambda_{q})\cdot\prod_{k=1,k\neq p}^{n}(\lambda_{k}-\lambda_{p})\prod_{l=1,l\neq q}^{n}(\lambda_{l}-\lambda_{q})}=\frac{P(r)}{Q(r)}$$

just reduce this summation to a common denominator which should be a fully squared form but hold a single negtive

$$Q(r)=-(\lambda_{r}-\lambda_{p})^2(\lambda_{r}-\lambda_{q})^2(\lambda_{p}-\lambda_{q})^2\cdot\prod_{m=1,m\neq p,q,r}^{m=n}\left((\lambda_{m}-\lambda_{p})^2(\lambda_{m}-\lambda_{q})^2\right)<0$$

and the numerator is a symmetric summation

$$P(r)=\sum_{p,q=1,p\neq q;p,q\neq r}^{p,q=n} {\left(\prod_{m=1,m\neq p,q,r}^{m=n}{(\lambda_{m}-\lambda_{p})(\lambda_{m}-\lambda_{q})}\right)}$$

in which, actually there should not be any $\lambda_{r}$, and then we only need to confirm that $P(r)$ is definitely positive. for the choose of $r$ is trivial in the proof, simply taking $r=1$, and here example the cases as $n=3,4,5$ ($n\ge3$)

for $n=3$, $P_{3}(1)=1$

for $n=4$, $$P_{4}(1)=\lambda_{2}^{2}+\lambda_{3}^{2}+\lambda_{4}^{2}-\lambda_{2}\lambda_{3}-\lambda_{2}\lambda_{4}-\lambda_{3}\lambda_{4}=\frac1{2}(\lambda_{2}-\lambda_{3})^2+\frac1{2}(\lambda_{2}-\lambda_{4})^2+\frac1{2}(\lambda_{3}-\lambda_{4})^2>0$$

for $n=5$, $$\begin{aligned} P_{5}(1) =&\;2\lambda_{2}^2\lambda_{3}^2 + 2\lambda_{2}^2\lambda_{4}^2 + 2\lambda_{2}^2\lambda_{5}^2 + 2\lambda_{3}^2\lambda_{4}^2 + 2\lambda_{3}^2\lambda_{5}^2 + 2\lambda_{4}^2\lambda_{5}^2\\ &-2\lambda_{2}^2(\lambda_{3}\lambda_{4}+\lambda_{3}\lambda_{5}+\lambda_{4}\lambda_{5})-2\lambda_{3}^2(\lambda_{2}\lambda_{4}+\lambda_{2}\lambda_{5}+\lambda_{4}\lambda_{5})\\ &-2\lambda_{4}^2(\lambda_{2}\lambda_{3}+\lambda_{2}\lambda_{5}+\lambda_{3}\lambda_{5})-2\lambda_{5}^2(\lambda_{2}\lambda_{3}+\lambda_{2}\lambda_{4}+\lambda_{3}\lambda_{4})+12\lambda_{2}\lambda_{3}\lambda_{4}\lambda_{5}\\ =&\;(\lambda_{2}\lambda_{3}-\lambda_{2}\lambda_{4})^2+(\lambda_{2}\lambda_{3}-\lambda_{2}\lambda_{5})^2+(\lambda_{2}\lambda_{4}-\lambda_{2}\lambda_{5})^2\\ &+(\lambda_{3}\lambda_{2}-\lambda_{3}\lambda_{4})^2+(\lambda_{3}\lambda_{2}-\lambda_{3}\lambda_{5})^2+(\lambda_{3}\lambda_{4}-\lambda_{3}\lambda_{5})^2\\ &+(\lambda_{4}\lambda_{2}-\lambda_{4}\lambda_{3})^2+(\lambda_{4}\lambda_{2}-\lambda_{4}\lambda_{5})^2+(\lambda_{4}\lambda_{3}-\lambda_{4}\lambda_{5})^2\\ &+(\lambda_{5}\lambda_{2}-\lambda_{5}\lambda_{3})^2+(\lambda_{5}\lambda_{2}-\lambda_{5}\lambda_{4})^2+(\lambda_{5}\lambda_{3}-\lambda_{5}\lambda_{4})^2\\ &-2(\lambda_{2}\lambda_{3}-\lambda_{4}\lambda_{5})^2-2(\lambda_{2}\lambda_{4}-\lambda_{3}\lambda_{5})^2-2(\lambda_{3}\lambda_{4}-\lambda_{2}\lambda_{5})^2 \end{aligned}$$

recalling

$$(a-b)^2+(b-c)^2+(c-d)^2+(d-a)^2\ge(a-c)^2+(b-d)^2$$

due to $(a+c)^2+(b+d)^2\ge2(a+c)(b+d)$

$$\begin{aligned} &\;(\lambda_{2}\lambda_{3}-\lambda_{2}\lambda_{4})^2+(\lambda_{4}\lambda_{2}-\lambda_{4}\lambda_{5})^2+(\lambda_{5}\lambda_{4}-\lambda_{5}\lambda_{3})^2+(\lambda_{3}\lambda_{5}-\lambda_{3}\lambda_{2})^2\\ \ge&\;(\lambda_{2}\lambda_{3}-\lambda_{4}\lambda_{5})^2+(\lambda_{2}\lambda_{4}-\lambda_{3}\lambda_{5})^2 \end{aligned}$$

and all of its symmetric analog, $P_{5}(1)>0$ can be proved as well. moreover, by this unscramble we also can write

$$P_{5}(1)=(\lambda_{2}-\lambda_{3})^2(\lambda_{4}-\lambda_{5})^2+(\lambda_{2}-\lambda_{4})^2(\lambda_{3}-\lambda_{5})^2+(\lambda_{2}-\lambda_{5})^2(\lambda_{3}-\lambda_{4})^2>0$$

so, the essential is how to write a complete generation for $P_{n}(r)$, I think $P_{5}$ is an enough instance for a good guess, where I may finish later.