# trace inequality

 Importance: Medium ✭✭
 Author(s):
 Subject: Algebra
 Keywords:
 Posted by: Miwa Lin on: October 11th, 2008

Let $A,B$ be positive semidefinite, by Jensen's inequality, it is easy to see $[tr(A^s+B^s)]^{\frac{1}{s}}\leq [tr(A^r+B^r)]^{\frac{1}{r}}$, whenever $s>r>0$.

What about the $tr(A^s+B^s)^{\frac{1}{s}}\leq tr(A^r+B^r)^{\frac{1}{r}}$, is it still valid?

% You may use many features of TeX, such as % arbitrary math (between $...$ and $$...$$) % \begin{theorem}...\end{theorem} environment, also works for question, problem, conjecture, ... % % Our special features: % Links to wikipedia: \Def {mathematics} or \Def[coloring]{Graph_coloring} % General web links: \href [The On-Line Encyclopedia of Integer Sequences]{http://www.research.att.com/~njas/sequences/}

## Bibliography

% Example: %*[B] Claude Berge, Farbung von Graphen, deren samtliche bzw. deren ungerade Kreise starr sind, Wiss. Z. Martin-Luther-Univ. Halle-Wittenberg Math.-Natur. Reihe 10 (1961), 114. % %[CRS] Maria Chudnovsky, Neil Robertson, Paul Seymour, Robin Thomas: \arxiv[The strong perfect graph theorem]{math.CO/0212070}, % Ann. of Math. (2) 164 (2006), no. 1, 51--229. \MRhref{MR2233847} % % (Put an empty line between individual entries)

* indicates original appearance(s) of problem.

### clarification

I suppose that A and B are square hermitian matrixes and that s and r are real numbers. But what are the brackets supposed to represent? Sorry for my ignorance if it is obvious.

It is open.

### this inequality is not true.

this inequality is not true. It was denied some years ago. See X.Z Zhan's