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6 votes
Accepted

Integral inequality: Prove $\int_0^1 f\int_0^1 1/f \leq 1$ for a certain function $f$

Matematika, shmatematika: any minimally decent CAS should immediately detect and tell the human operator that, for $r=1$, the inequality reads $$ (\int_X F-\int_Y G)(\int_X 1/F-\int_Y 1/G)\le 1 $$ ...
  • 54.4k
4 votes

Integral inequality: Prove $\int_0^1 f\int_0^1 1/f \leq 1$ for a certain function $f$

The inequality in question is equivalent to the following: $$L(X):=Ef(X)\,E\frac1{f(X)}\le1,$$ where $$f(x):=x\sqrt{1-r/x^2}$$ and $X$ is a random variable (r.v.) such that $X^2>r$ and $EX=0$. By ...
12 votes

For what kind of $C^*$ algebras does the inequality $\frac{(ab+ba)}{2}\leq\frac{ a^p}{p} +\frac{b^q }{q}$ hold for $a,b>0$?

Let me expand slightly on the comments I made above, and give the most general solution. Clearly the inequality $\frac{ab + ba}{2} \leq \frac{a^2}{2} + \frac{b^2}{2}$ holds for all positive elements $...
  • 2,151
3 votes

Proof of a matrix implication

Using Groebner bases and inequalities on $\mathbb{R}$, we prove that the OP's conjecture is true. The real polynomial $x^2-sx+p$ admits roots of modulus $<1$ iff $p<1,p>-1,p>-s-1,p>s-1,...
  • 2,968
2 votes

Proof of a matrix implication

By considering cases of real/complex eigenvalues, the problem is reduced to a system of polynomial inequalities in each of the four cases. For example, the case of real eigenvalues of $A^2B$ and non-...
1 vote

Is there an inequality relation between KL-divergence and $L_2$ norm?

This is probably obvious, but just wanted to explicitly mention the following. It is perhaps more natural to look at the modified $L_2$ norm $L_2'(p,q) = L_2(\sqrt{p}, \sqrt{q})$. Note that ...
  • 1,033
1 vote

Elementary convexity example

Here is a much simpler proof, actually of the more general fact that $$f(x):=x^p(1+\ln^+ x)^s$$ is convex in $x\ge0$ for any real $p\ge1$ and $s\ge0$, where $\ln^+ x:=\ln\max(1,x)$. For the left and ...
4 votes
Accepted

On existence of a concave function

Such a function doesn't exist for some choices of $a$. For notational purposes I will change the unit interval by $[-2,2]$. Consider a $C^\infty$ function $a:[-2,2]\to\mathbb{R}$ such that $a(0)=3$, $...
  • 5,377
6 votes
Accepted

Inequality with decreasing rearrangement function

No (if $c$ cannot depend on $f^*$ or $g$). Indeed, let $h:=f^*/g$. Then $h$ can be any positive function and the inequality in question can be rewritten as $$lhs:=\Big(\int_0^\infty h(s)^{p'}ds\Big)^{...
8 votes

How to prove that $1/ ((y+z) x^4) + 1/ ((z+x) y^4) + 1/ ((x+y) z^4) \geq 3/2$ for $x, y, z>0$ such that $xyz=1$?

Is the "cauchy-schwarz-inequality" tag a guess or a hint? . . . At any rate, it turns out to be a good start. Let $$ R := \frac1{(y+z) x^4} + \frac1{(z+x) y^4} + \frac1{(x+y) z^4}. $$ We ...
8 votes

How to prove that $1/ ((y+z) x^4) + 1/ ((z+x) y^4) + 1/ ((x+y) z^4) \geq 3/2$ for $x, y, z>0$ such that $xyz=1$?

$\newcommand\tH{\tilde H}$This problem is one of real algebraic geometry, which can be solved purely algorithmically. In Mathematica, such algorithms are implemented by ...
6 votes

How was Claim 5 in "A non-linear generalisation of the Loomis–Whitney inequality and applications" thought up?

As Dan Romik points out, this technique is relatively old folklore by now. Terry Tao calls this the "tensor power trick" in a blog post dedicated to the subject; the two elementary ...
6 votes

How was Claim 5 in "A non-linear generalisation of the Loomis–Whitney inequality and applications" thought up?

An instance of this idea of killing an unwanted factor in an inequality by considering an inequality for $k$-th powers and then taking the limit as $k\to\infty$ appears in the proof of the Kraft-...
  • 2,226
2 votes

Is there an inequality relation between KL-divergence and $L_2$ norm?

Now, I am trying to answer this question. Proposition. If $p$ and $q$ are two probability densities, and (upper) bounded by $\tau_1$ and $\tau_2$, respectively, then $$ KL(p,q) \ge \frac{1-\log(2)}{\...
  • 165
1 vote

Finding an analytical upper bound on linear transform of matrix

A result assuming that $M$ is positive definite. By continuity, in the optimal $\alpha$ the matrix $M+\alpha D$ is singular; hence the result is the smallest zero of $f(\alpha) = \det (M + \alpha D)$, ...
11 votes
Accepted

a problem in complex-variable inequality

For a polynomial $Q(x)=\sum_i q_ix^i$, define $N(Q)=\sum_i|q_i|^2$. We need to show that $N(R)\geq 2$, where $R(x)=\prod_i (x-z_i)$. For a polynomial $Q(x)=\sum_{i=0}^k q_ix^i$, define $Q^*(x)=\sum_{i=...

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