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I really appreciate it when you take time to answer my questions. Thanks!


Jan
22
comment Inequality: $(a^3+a+1)(b^3+b+1)(c^3+c+1) \leq 27$
See Shane Chern's answer.
Jan
22
comment Inequality: $(a^3+a+1)(b^3+b+1)(c^3+c+1) \leq 27$
This inequality can be solved by noting that $x \to \ln(x^{3/2}+x^{1/2}+1)$ is concave for $x > 0$ and Jensen's inequality. Since dineshdileep noticed something similar first, I won't post this as answer. (Unless he doesn't edit his answer)
Jan
22
comment Inequality: $(a^3+a+1)(b^3+b+1)(c^3+c+1) \leq 27$
my bad! I should have done a change of variable. In fact $x \to \log(x^{3/2}+x^{1/2}+1)$ is concave on $x > 0$. I am not familiar with the terminology of Schur concavity, but usual Jensen's inequality would work.
Jan
22
comment Inequality: $(a^3+a+1)(b^3+b+1)(c^3+c+1) \leq 27$
In fact, no, it can't be Schur-convex/concave, because $log(x^3+x+1)$ is neither convex or concave. This means that one can cook up counterexamples by picking points in the region where the function is convex then picking points where the function is concave.
Jan
22
comment Inequality: $(a^3+a+1)(b^3+b+1)(c^3+c+1) \leq 27$
Product of convex functions may not be convex. In fact LHS is neither convex or concave on $[0,3]$. I'm not sure about Schur-convexity/concavity, but I don't think it's something you can handwave.
Jan
22
comment Inequality: $(a^3+a+1)(b^3+b+1)(c^3+c+1) \leq 27$
How does monotonically increasing imply concavity? When one vector majorizes another, it's possible that some components are larger, while some components are smaller.
Jan
22
comment Inequality: $(a^3+a+1)(b^3+b+1)(c^3+c+1) \leq 27$
I think you mean it's Schur concave. Why is it true though?
Jan
21
comment Detail about the definition of orientability
@M.Luethi, yes.
Jan
21
comment Detail about the definition of orientability
@M.Luethi, I'm pretty certain that this works.
Jan
21
comment Detail about the definition of orientability
2. Non-vanishing global sections may not exist. How do you deal with the zeros of sections?
Jan
21
comment Detail about the definition of orientability
I still don't quite understand it. 1. You seem to be connecting sections. Can you make it a bit more precise how this connects to the connectedness of the line bundle - zero section?
Jan
21
comment Detail about the definition of orientability
What do you mean by for all $\omega \neq 0$ you would get path-connectedness?
Jan
21
comment Show that satisfaction of Cauchy-Riemann Equations in polar coordinates implies analyticity
CR-equations in rectangular/polar coordinates are really the same equation except that they are in different coordinates. So the result for rectangular coordinates already imply that for polar coordinates.
Jan
21
comment Inequality involving partial sums of $\frac{|\sin{kx}|}{k}$
Can you say more about "filling in a few details"? It's very unclear to me how you may deal with the problem of different phase.
Jan
20
comment How to proof M is a compact connected n-manifold,if for any point $p\in M,M\backslash\{p\}\cong R^n$ then M is homeomorphic to $S^n$.
See ncatlab.org/nlab/show/one-point+compactification . Your friend's hint is about the uniqueness of one point compactification. See the third property of my link here.
Jan
20
comment Uniqueness of Hermitian inner product
@K.Ghosh, try to work out how $G$ acts on $Hom(V, \bar{V})$, and what it means for a map to be $G$-invariant.
Jan
20
comment $\dfrac{(b+c-a)^2}{(b+c)^2+a^2}+ \frac{(c+a-b)^2}{(c+a)^2+b^2}+ \frac{(a+b-c)^2}{(a+b)^2+c^2} \ge \frac{3}{5}$
@harrypham, if not, replace $a$ by $\frac{3a}{a+b+c}$ and so forth. Left hand side doesn't change this way.
Jan
20
comment $\dfrac{(b+c-a)^2}{(b+c)^2+a^2}+ \frac{(c+a-b)^2}{(c+a)^2+b^2}+ \frac{(a+b-c)^2}{(a+b)^2+c^2} \ge \frac{3}{5}$
@Marvis, You can check by clearing the denominator that it is equivalent to $(2a+1)(a-1)^2 \ge 0$. As for the motivation, let $f(a) = \frac{(3-2a)^2}{(3-a)^2+a^2}$. Then $f(1) = \frac{1}{5}$, and $f'(1) = -\frac{18}{25}$.
Jan
20
comment a neighbourhood of identity U generates G where g is a connected lie group
@K.Ghosh, I believe so.
Jan
19
comment Homotopy equivalence
@Anne, do you know van Kampen's theorem?