## New answers tagged inequalities

2

Using (say) the Jordan form, approximate $A$ by a nonsingular matrix $C_t$ so that $C_t\to A$ as $t\to0$. Then
$$C_t^\top (C_t X^{-1} C_t^\top + B)^{-1} C_t\le
C_t^\top (C_t X^{-1} C_t^\top)^{-1} C_t=X.$$
Letting now $t\to0$, we get
$$A^\top (A X^{-1} A^\top + B)^{-1} A \le X,$$
as desired.

14

Yes, because the OP stated that the ground field is $\mathbb{R}$, one can simply take the octic polynomial
$$
Q(v_1,v_2,\ldots,v_n) = \sum_{1\le i < j\le n} \bigl((v_i,v_i)(v_j,v_j)-(v_i,v_j)^2\bigr)^2,
$$
which will do the trick.

4

An affirmative answer to your question would follow from a famous conjecture about envy-free allocations. It is also known that if you replace the '2' in $b$-goodness with '$\frac 12$', then it is true.
A good paper to read (first) on the topic is Almost Envy-Freeness with General Valuations.
An allocation of some items among some players is envy-free up to ...

0

Just for those who might want to know, I think the min-max theorem mentioned by @Michael Renardy is Courant-Fischer: Supporse a real symmetric matrix A's eigen values are $\lambda_1 \leq \lambda_2 \leq ... \leq \lambda_n $. then k-th eigen value: $$ \lambda_k = \underset{V_k}{\text{min}} \space max\{x^TAx | x \in V_k, ||x||_2 = 1\}$$
$V_k$ is any k-...

3

I was recently thinking about this question again and I came up with another way to explain the result $$\mathcal H_{\lambda}=\left\{ h(i,j)\right\}_{(i,j)\in \lambda}\succcurlyeq \left\{ h^{*}(i,j)\right\}_{(i,j)\in \lambda}=\mathcal H^{*}_{\lambda} \tag{1}$$
in a way that (i) combinatorially describes the sequence of Robin Hood moves to get from $\mathcal ...

0

Regarding the inequality for $\Vert AB \Vert_2 $, although it does not hold in the form you wrote it as Denis pointed out, replacing the 2-norm by the 1-norm leads to Holder's inequality for Schatten norms:
\begin{equation}
\Vert AB \Vert_1 \leq \Vert A \Vert _p \Vert B \Vert _q
\end{equation}

10

The inequality is false for $a=1799, b=105, c=1024, d=4$.
This counterexample was found and verified with Mathematica, as follows:

1

Weyl's inequality, and Hua's inequality.
They are quite important from the viewpoint of analytic number theory.

5

I have nothing to add to the Fourier-type approach suggested in the question, but for those curious, thought it useful to outline the combinatorial solution to the problem that I know (I believe this is the same as the IMO official solution, and claim no originality).
One thing to add is that, although induction is a crucial part of the proof, we do not use ...

0

This can be done with help of Maple in such a way. First, we find the estimated expression explicitly by
a := (exp(x)*int(sin(exp(t)), t = x .. x + 1) assuming x::real;
$ {{\rm e}^{x}} \left( -{\it Si} \left( {{\rm e}^{x}} \right) +{\it Si}
\left( {{\rm e}^{x+1}} \right) \right)
$
In fact, the integral is reduced to another integrals. Next, the ...

2

We have $0\le a\le b\le1$ and $T\in(0,\infty)$. We want to know when there is a positive constant $C$ such that
$$\int_0^T dt\, \int_a^b dx\, u^2(x-t)\geq C\int_0^1 dx\,u^2(x) \tag{1}$$
for all measurable functions $u\colon\mathbb R\to\mathbb R$ such that $u(x)=0$ for $x\notin(a,b)$.
The answer is: never. Indeed, without loss of generality $a<b$. The left-...

6

For any unit vector $x\in L^2(\mathbb{R}^n)$,
$$\|\sqrt{A+B}\,Cx\|^2=(\sqrt{A+B}\,Cx,\sqrt{A+B}\,Cx)
=(ACx,Cx)+(BCx,Cx)=\|\sqrt A\,Cx\|^2+\|\sqrt B\,Cx\|^2
\le(\|\sqrt A\,Cx\|^2+\|\sqrt B\,Cx\|)^2
\le(\|\sqrt A\,C\|+\|\sqrt B\,C\|)^2.$$
So,
$$\|\sqrt{A+B}\,C\|\le\|\sqrt A\,C\|+\|\sqrt B\,C\|.$$

8

$$
\|\sqrt{A+B}Cx\|^2=(\sqrt{A+B}Cx,\sqrt{A+B}Cx)=\\
((A+B)Cx,Cx)=(ACx,Cx)+(BCx,Cx)=\|\sqrt{A}Cx\|^2+\|\sqrt{B}Cx\|^2,
$$
taking the supremum over unit vectors $x$ we get
$$
\|\sqrt{A+B}C\|^2\leqslant \|\sqrt{A}C\|^2+\|\sqrt{B}C\|^2\leqslant (\|\sqrt{A}C\|+\|\sqrt{B}C\|)^2.
$$

4

For any real numbers $u,v,c$ such that $u\le c\le v$, let $\mu_{c;u,v}$ denote the unique probability distribution on the set $\{u,v\}$ with mean $c$.
Your generalization of Jensen's inequality follows immediately from the well-known fact that any probability distribution $\mu$ on $\mathbb R$ with a given mean $c\in\mathbb R$ is a mixture of probability ...

0

$C=1.0986702957852257$ is the solution. It can be proved by using the Langrange multipliers.

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