polynomial with real coefficients

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Source: IMC 1998 day 1 problem 5

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Peter

#1 h Nov 1, 2005, 12:47 AM • 2 Y

Y by Adventure10, Mango247

Let $P$ be a polynomial of degree $n$ with only real zeros and real coefficients.
Prove that for every real $x$ we have $(n-1)(P'(x))^2\ge nP(x)P''(x)$ . When does equality occur?

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#2 h Nov 26, 2005, 10:42 AM • 3 Y

Y by dinamometre1123, Adventure10, Mango247

I will proceed by induction. For $n=1$ that's manifestly true. Let $P$ be polynomial of degree $n+1$ , so
$P(x)=u(x)Q(x)$ where $u(x)$ is a monic real polynomial of degree $1$ and $Q(x)$ is a polynomial of degree $n$ with only real roots and real coefficients.
$P(x)=u(x)Q(x) \Longrightarrow P'= Q+uQ'\Longrightarrow P'' = 2Q' + uQ''$ , where $(n-1)(Q')^2\geq nQQ''$ , so :

$n(P')^2-(n+1)PP'' = nQ^2-2uQQ'+nu^2(Q')^2-(n+1)u^2QQ''\geq$ $nQ^2-2uQQ' + \frac{u^2(Q')^2}n = (Q\sqrt n - \frac{uQ'}{\sqrt n})^2 \geq 0$ ... (q.e.d)

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#3 h Nov 26, 2005, 11:03 AM • 2 Y

Y by Adventure10, Mango247

Sorry, I forgot the equality case ! Remember : $u(x)$ is a monic real polynomial , $u(x) = x-a$
$n(P')^2-(n+1)PP'' = 0 \Longrightarrow Q\sqrt n - \frac{uQ'}{\sqrt n} = 0 \Longrightarrow \frac{Q'}{Q} = \frac nu$ $\Longrightarrow Q(x)= cu^n_{(x)}\Longrightarrow P(x)=c u^{n+1}_{(x)}$
Also, It's easy to verify that the equality holds for $P(x)=c u^{n+1}_{(x)}$

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#4 h Aug 11, 2009, 11:33 PM • 2 Y

Y by Davrbek, Adventure10

without induction, by the condition we have that
$P = b\prod^n_{i=1} (x-a_i)$
Suppose $x$ is not a root of $P$ and Let $c_i = \frac{1}{x-a_i}$ , then
$P' = P (c_1 + c_2 + \ldots + c_n), P'' = P (\sum_{i \neq j} 2c_i c_j)$
The problem thus becomes
$(n-1)(c_1 + c_2 + \ldots + c_n)^2 \ge 2n (\sum_{i \neq j}c_i c_j)$
factoring gives
$\sum_{i \neq j} (c_i - c_j)^2 \ge 0$
Which is trivially true. Also the equality holds iff all the $c_i$ are equal, which means all the $a_i$ are equal and $P=b(x-a)^n$

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chakrabortyahan

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chakrabortyahan

#5 h Apr 5, 2024, 8:43 PM

Y by

Another solution
Let $\alpha_1,\alpha_2,...\alpha_n$ be the zeroes (non-necessarily distinct) of $P$ . The given inequality holds trivially for $\alpha_i's$ . So we assume that $x\neq \alpha_i$ in the latter part of the solution.
Now note that $P'(x) = P(x) [\sum_{i=1}^n\frac{1}{x-\alpha_i}]$
Now by C-S inequality ,
$n\cdot\sum\limits_{i=1}^n\frac{1}{(x-\alpha_i)^2}\ge \left[ \sum\limits_{i=1}^n\frac{1}{|x-\alpha_i|}\right]^2\ge \left[\sum\limits_{i=1}^n\frac{1}{x-\alpha_i}\right]^2...(i)$
Now note that $(i)$ can be re-written as $-n\left[\frac{P'(x)}{P(x)}\right]'\ge \left[\frac{P'(x)}{P(x)}\right]^2\iff 0\ge \frac{n(P''(x)P(x)-P'(x)^2)}{P(x)^2}+\frac{P'(x)^2}{P(x)^2}$
Hence proved as $P(x)^2> 0$ for all $x\neq \alpha_i$
And the equality holds iff all $\alpha_i$ s are equal (equality case of Cauchy -Schewarz)
$\blacksquare\smiley$

This post has been edited 1 time. Last edited by chakrabortyahan, May 5, 2024, 6:22 PM

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#6 h Apr 5, 2024, 10:09 PM

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This problem was used for oral examination ENS PSLR in 2023
problem 23 oral PSLR
https://www.rms-math.com/exos-etoiles-2023-site.pdf

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#7 h Mar 24, 2025, 11:39 AM

Y by

chakrabortyahan wrote:

Another solution
Let $\alpha_1,\alpha_2,...\alpha_n$ be the zeroes (non-necessarily distinct) of $P$ . The given inequality holds trivially for $\alpha_i's$ . So we assume that $x\neq \alpha_i$ in the latter part of the solution.
Now note that $P'(x) = P(x) [\sum_{i=1}^n\frac{1}{x-\alpha_i}]$
Now by C-S inequality ,
$n\cdot\sum\limits_{i=1}^n\frac{1}{(x-\alpha_i)^2}\ge \left[ \sum\limits_{i=1}^n\frac{1}{|x-\alpha_i|}\right]^2\ge \left[\sum\limits_{i=1}^n\frac{1}{x-\alpha_i}\right]^2...(i)$
Now note that $(i)$ can be re-written as $-n\left[\frac{P'(x)}{P(x)}\right]'\ge \left[\frac{P'(x)}{P(x)}\right]^2\iff 0\ge \frac{n(P''(x)P(x)-P'(x)^2)}{P(x)^2}+\frac{P'(x)^2}{P(x)^2}$
Hence proved as $P(x)^2> 0$ for all $x\neq \alpha_i$
And the equality holds iff all $\alpha_i$ s are equal (equality case of Cauchy -Schewarz)
$\blacksquare\smiley$

$n\cdot\sum\limits_{i=1}^n\frac{1}{(x-\alpha_i)^2}\ge \left[ \sum\limits_{i=1}^n\frac{1}{|x-\alpha_i|}\right]^2\ge \left[\sum\limits_{i=1}^n\frac{1}{x-\alpha_i}\right]^2...(i)$
why did you have modulus here in the thing between? CS inequality is true for all reals right

This post has been edited 1 time. Last edited by mqoi_KOLA, Mar 24, 2025, 11:39 AM

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#8 h Mar 26, 2025, 5:00 PM

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@above $\alpha_i$ can be complex ig?

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