$\prod_{k=1}^{n}(x_k-a)(x_k-b)\leqslant\sum_{q=1}^{n}x_q^2\prod_{p=1,p\neq q}^{n}(x_p-a)(x_p-b).$?

Question

Is there a name for this formula? For $f_k,w_k\geqslant0$.

$$\prod_{k=1}^{n}f_k\leqslant\sum_{q=1}^{n}w_q\prod_{p=1,p\neq q}^{n}f_p.$$

I believe that there is $w_k$ that make the formula true. Am I right?

EDIT: $f_k=(x_k-a)(x_k-b)$ and $w_k=x_k^2$, $a$ and $b$ are the roots of $f_k$.

So I want to prove this:

$$\prod_{k=1}^{n}(x_k-a)(x_k-b)\leqslant\sum_{q=1}^{n}x_q^2\prod_{p=1,p\neq q}^{n}(x_p-a)(x_p-b).$$

you need to add additional context. if you want it to hold for general polynomials $f_n$ and some constants $w_q$ then it's not gonna work. if you take all $f_n$'s to be just $f(x) = x$ then the LHS is degree $n$ while RHS is of degree $\leq n-1$ so that can't hold. — mm-aops, Jun 10 '14 at 22:03
By considering the ratio RHS/LHS, you just want to prove: $$\sum_{q=1}^{n}\frac{x_q^2}{(x_q-a)(x_q-b)}\geq 1,$$ but this inequality cannot hold if all the $x_q$s are located between $a$ and $b$, for instance. — Jack D'Aurizio, Jun 14 '14 at 16:08
Given the similarity with this other recent question where the roots are complex, can you state the reference? — Calvin Lin, Jun 15 '14 at 16:16

score 0 · Accepted Answer · answered Jun 13 '14 at 17:06

In general that inequality does not hold.

Taking $x_1=a$ and $x_2=\cdots=x_{n-1}=0$ for example, the inequality becomes $(ab)^{n-2}(x_n-a)(x_n-b) \ge 0$, which is clearly not true in general. If $a\neq b$ then the sign of $(x_n-a)(x_n-b)$ can change (from $-\infty$ to $+\infty$, the order is $+-+$).

$­\prod_{k=1}^{n}(x_k-a)(x_k-b)\leqslant\sum_{q=1}^{n}x_q^2\prod_{p=1,p\neq q}^{n}(x_p-a)(x_p-b).$?

1 Answers1

$\prod_{k=1}^{n}(x_k-a)(x_k-b)\leqslant\sum_{q=1}^{n}x_q^2\prod_{p=1,p\neq q}^{n}(x_p-a)(x_p-b).$?