## New answers tagged inequalities

2

The inequality does not always hold. Let $D \sim$ Bernoulli$(p)$, independent of $Y$. Let $Y'$ and $Y''$ be two independent copies of $Y$ (also independent of $Y$). Let $X=D Y+(1-D) Y'$ and $Z=DY'' + (1-D)Y$. Define also $m_k=E(Y^k)$ for $k=1,2$. Then it is easy to check that Cov$(X,Y) = p V(Y)$ and Cov$(Z,Y) = (1-p) V(Y)$. Because $X$ and $Z$ have the same ...

2

$\newcommand{\tnu}{\tilde\nu}$Continuing Alex Ravsky's comment, we have
\begin{equation*}
2^{2^n}\tnu_{n,1}=\sum_{j=0}^{2^n-1}S_j=S_0+(2^n-1)S_1, \tag{1}
\end{equation*}
where
\begin{equation*}
S_j:=\sum_{F\subseteq[2^n]}\Big|\sum_{i\in F}a_{i,j}\Big|,
\end{equation*}
and for each $j\ne0$
\begin{equation*}
S_j=S_1=\sum_{k,l=0}^M\binom Mk \binom ...

1

Certainly not, if $C$ is supposed to be real. For instance, suppose that $\rho_1$ and $\rho_2$ are probability densities such that $\rho_1\rho_2=0$. Then the left-hand side of your inequality is $\infty$, whereas its right-hand side is $2C$.

1

Yes: by the generalized mean inequality (or, more specifically, by the AM--QM inequality), $\sqrt{nS}$ is an upper bound on $\sum_{k=1}^n\sqrt{s_k}$, which is better than
$n\sqrt{\max_{1\le k\le n}s_k}$.

1

By scaling if necessary, we may assume without loss of generality that $a^2p + b^2\bar p = 1$ Substituting $u = a^2p$, we can rewrite the 1-dimensional inequality as
\begin{align*}
f(u) := u\log \frac{u}{p} + \bar u \log \frac{\bar u}{\bar p} - c(p) (\sqrt{\bar u p} - \sqrt{u \bar p})^2 \le 0.
\end{align*}
We calculate the first two derivatives of $f(u)$. ...

3

This inequality is known. Indeed, inequality (1.1) by Kwapień and Szulga states that
$$\Big(\int_S\Big(\int_T|f(s,t)|^q\mu(dt)\Big)^{p/q}\nu(ds)\Big)^{1/p} \\
\le
\Big(\int_T\Big(\int_S|f(s,t)|^p\nu(ds)\Big)^{q/p})\mu(dt)\Big)^{1/q}$$
if $0<q<p$, where $\mu$ and $\nu$ are measures.
The case $p<q<0$ is obtained from this by replacing $f$ (which ...

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