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 May 2 comment Is $\int_{\mathbb R} f(\sum_{k=1}^n\frac{1}{x-x_k})dx$ independent of $x_k$'s for certain $f$? I have noticed that Theorem 3 in sos440's answer to the original post about this topic has solved the problem in my last comment. Thank you! May 2 comment Is $\int_{\mathbb R} f(\sum_{k=1}^n\frac{1}{x-x_k})dx$ independent of $x_k$'s for certain $f$? Dear David Speyer, thank you for your excellent answer! I still have one question. The displayed equality in your answer also seems to hold for some $f$ that fails to be holomorphic somewhere, say $f(z)=\frac{1}{1+|z|^\alpha}$ for real $\alpha>2$. Could such $f$ be covered as a simple corollary of your result? Apr 29 comment Is $\int_{\mathbb R} f(\sum_{k=1}^n\frac{1}{x-x_k})dx$ independent of $x_k$'s for certain $f$? @DavidSpeyer: Thank you for your comments. I agree with your observation. I think in fact the statement for general positive real $a_k$ can be deduced from the statement for integers. First, by change of variables for $x$, the statement holds for rational $a_k$; second, use rationals to approximate reals. Apr 29 comment Is $\int_{\mathbb R} f(\sum_{k=1}^n\frac{1}{x-x_k})dx$ independent of $x_k$'s for certain $f$? @SamratMukhopadhyay: Thank you, but I have no idea how to apply argument principle for this problem. Apr 29 comment On the inequality $\int_{-\infty}^{+\infty}\frac{(p'(x))^2}{(p'(x))^2+(p(x))^2}\,dx \le n^{3/2}\pi.$ @DavidSpeyer: I have posted the problem. Please refer to the linked question to this post. Apr 29 comment On the inequality $\int_{-\infty}^{+\infty}\frac{(p'(x))^2}{(p'(x))^2+(p(x))^2}\,dx \le n^{3/2}\pi.$ @DavidSpeyer: You are welcome. Yes, I also got stuck in showing the independence of $x_k$. It seems to work for many $f$'s; for example $f(t)=\frac{t^{2m}}{1+t^{2n}}$, $1\le m\le n$ are integers. Would you mind if I post a new question for this problem? Apr 29 comment On the inequality $\int_{-\infty}^{+\infty}\frac{(p'(x))^2}{(p'(x))^2+(p(x))^2}\,dx \le n^{3/2}\pi.$ @DavidSpeyer: The statement in your last comment is equivalent to $\int f(\sum_{k=1}^n\frac{1}{x-x_k})dx=n\cdot\int f(\frac{1}{x})dx$ for $f(t)=\frac{t^2}{1+t^2}$. It seems that the same equality holds for more general $f$, but I am completely clueless about the reason. Jan 28 comment Order of Double Coset @Sushil: The natural 1-1 and onto map is $axb\leftrightarrow axbx^{-1}$ for $a\in A$ and $b\in B$. Jan 6 comment Annoying Polynomial Inequality @SouvikDey: Please note that: (1) $h(x)=0$ if and only if $p(x)=0$; (2) $p$ is a polynomial. Dec 27 comment How to show that $f'(x)<2f(x)$ @Hans: Sorry, I don't know any other similar question, and after some failed attempts, I have to admit that I cannot explain my motivation of the "by (1)" step better than the post itself suggests. Dec 25 comment Holomorphic function $\varphi$ with fixed point $z_0$ such that $\varphi'(z_o)=1$ is linear? @VictorWang: You are welcome. Dec 25 comment Holomorphic function $\varphi$ with fixed point $z_0$ such that $\varphi'(z_o)=1$ is linear? @VictorWang: Sorry, I just read your comment because it is very long time since the last time I visited this website. Please note that since $\varphi(\Omega)\subset \Omega$, $\varphi^k$ is a well-defined holomorphic function on $\Omega$ for every $k$. Then from the choice of $r$ we know the radius of convergence of the Taylor expansion of $\varphi^k$ around $z_0$ is no less than $r$. Dec 25 comment How to show that $f'(x)<2f(x)$ @Hans: Thank you for your upvote. I think the last time I visited this website is a couple of months ago, and I just read your comment. I am not sure how to explain the motivation of the step mentioned in your comment clearly. Are you still interested in this question? Apr 4 comment Vector spaces and intersections @BeniBogosel: Thank you. Maybe I should mention that the example is a special case of the following observation. By choosing the basis $\{e_1,\dots, e_4\}$ appropriately, $V_1$, $V_2$, $V_3$ can always be of the form given in my answer and at the same time, $V_4$ is spanned by a pair of vectors $(e_1,e_3)A+(e_2,e_4)B$, where $A$, $B$ and $A-B$ are invertible $2\times 2$ matrices. Then $W$ exists iff $A^{-1}B$ has a real eigenvalue. In my answer, $A$ is identity and $B$ is rotation by $-\frac{\pi}{2}$. Mar 10 comment A beautiful inequality for convex functions @Julien: You are welcome! Mar 10 comment A beautiful inequality for convex functions @Julien: I edited my answer by reorganizing the argument. I hope it looks clearer now. In the current form, your question asks how I get $h_t$ in $(1)$. First I assume $h_t(x)=b\cdot V(x-a)$ for some constants $a$, $b$. Then from $h_t\ge f_t$ we know $b(t-a)=h_t(t)\ge f_t(t)=1$. To get $\int_0^1h_t \le 2\int_0^1f_t=2(1-t)$, I use a stronger inequality $\frac{b(1-a)^2}{2}=\int_a^1 h_t\le 2(1-t)$ instead. The inequalities determine a unique pair of $(a,b)=(2t-1,\frac{1}{1-t})$. There is probably more geometrical explanation, but maybe it needs more words due to the limitation of my English. Mar 9 comment A beautiful inequality for convex functions @Julien: I didn't show the motivation of the construction of $g$ in the linked answer, but the basic idea is quite similar to your approach, and the same idea can be applied to constructing $h$. Do I need to post an answer here to show all the details(probably it does not satisfy the "more intuitive" requirement in the earlier version of your question)? Mar 9 comment A beautiful inequality for convex functions @Julien: To see the history of all the editions of a post, you may simply click "edited +time" in the last line of the post. Mar 9 comment A beautiful inequality for convex functions @Julien: I almost forgot all the details in my answer to the linked question, and I just looked back at it briefly. It seems that the same argument there does not work, because it uses the fact that the supremum of convex functions is still convex, which fails to be true for infimum in general. However, I think the method in the remark of my answer there(or the first version of my answer) still works for constructing $h$, and this method shares the same basic idea with your own approach. Mar 9 comment A beautiful inequality for convex functions Here is a closely related question.