Consider a polynomial $p \in \mathbb{R}[x]$ of degree $n$ and with no real roots. Prove that$$\int_{-\infty}^{\infty}\frac{(p'(x))^2}{(p(x))^2+(p'(x))^2}dx$$converges, and is less or equal than $n^{3/2}\pi$ My approach Now let $x_1, x_2, \dots, x_n$ be the roots of $p$. By Cauchy-Schwarz $$(\sum_{k=1}^{n}{\frac{1}{x-x_k}})^2\leq n\sum_{k=1}^{n}{\frac{1}{|x-x_k|^2}}$$ I don't know what to do next. If I am wrong kindly provide a detailed answer in the answer section. I have shown what I have thought of or what I have done . Can anyone confirm if my thought process is right? Just a reminder... This question has been lying unanswered for a long time Thank you.
$ p(x)\in\mathbb{R[X]} $ is a polynomial of degree $n$ with no real   roots. Show that: $$\int\limits_{-\infty}^{+\infty}\dfrac{(p'(x))^2}{(p'(x))^2+(p(x))^2}\,dx \leq n^{3/2}\pi.$$   It's easy to see that the degree of $ p$ has to be even. For $n=2$ this integral is at most $2\pi$. For $n&gt;2$ the maximum value of this integral is obtained when all the imaginary parts of the roots of $p(x)$ tend to $0$, but I couldn't go further.    Any help would be appreciated, thanks.    Edits by David Speyer: It seems very likely now that the optimum bound is $n \pi$, not $n^{3/2} \pi$. As pointed out in the comments below, and further in , this is the value we get in the limit where $p$ has double roots on the real axis.   It seems likely that  moving the roots of $p$ away from the real axis can only make the integral less. Write the roots of $p$ as $a_j \pm i b_j$, so $p(x) = \prod ((x-a_j)^2 + b_j^2)$ and  $$\frac{p'(x)}{p(x)} = \sum \frac{2 (x-a_j)}{(x-a_j)^2 + b_j^2}.\quad (\ast)$$  So making the $b_j$ larger tends to make $p'(x)/p(x)$ smaller, which makes the integral smaller. But this argument is not rigorous, because the terms of $(\ast)$ can have both positive and negative sign, so it could be that making the individual terms closer to $0$ makes the absolute value of $(\ast)$ larger.  I don't see how to beat this issue easily.  Thus, I'm putting up a bounty for proving or disproving   $$ \int_{-\infty}^{+\infty}\dfrac{(p'(x))^2}{(p'(x))^2+(p(x))^2}\,dx \leq n\pi.$$
I'm pretty sure that you can't take a magnifying glass and make it focus to a point that is hotter than the surface of your light source. For example, when you're outside trying to fry ants with your magnifying glass, it's impossible to get it hotter than 5,000 C (the temperature of the surface of the sun). My dad was arguing with me about it because he didn't see why this was true. My easiest argument was that the 2nd law of thermodynamics prevents this from happening because heat can't flow passively from a place of lower energy/entropy to a place of higher energy/entropy. He didn't buy it, saying that there wasn't anything preventing the light from focusing to a hotter point.  So I was wondering, are there some laws of optics that prevent this from happening? Or alternatively, is there a way to show that you could build a perpetual motion machine from this? Any help is appreciated. How can I change the font of the top bar without changing the shell theme? 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SE Android app for tablet has bad formatting on feed The lines wrap so much on v1.0.85 that only a few items can be displayed.  Edit: This is on the world's most low end tablet, Azpen A727.  Screenshots of the old:    and new, showing loss of quality My 13 year old daughter says she wants to be a boy. This is out of character for her. Is this a phase? My daughter, 13, has said she wants to be a boy. A little background: we moved to a new state a year and a half ago. In her new school, she felt isolated among her peers until she met a girl in art, and became good friends. She joined the drama club, and made  friends with some 8th graders. They joined “the diversity club”. This club is a LGBT club. The kids put on a play they wrote and starred in, which had homosexual connotations. “Romeo and Juliet”, but script  rewritten to include  2 females as the couple, for instance.   My daughter got so wrapped up in it, that her life was as if she was copying the play. 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