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Can someone explain to me why the following does not simplify to True (as all constituents are positive)?

FullSimplify[a+b/(c+(d-e)^2)>0,Assumptions->And@@Thread[{a,b,c,d,e}>0]]

Most special cases including e=0 or a=b,... do simplify correctly.

What I eventually want to do is to simplify expressions of the kind 

Abs[a+b/(c+(d-e)^2)],

which should evaluate to a+b/(c+(d-e)^2) under given assumptions.

Any workarounds?

Edit
What I ended up doing for larger expressions was this:

expr[Abs[a+b/(c+(d-e)^2)]]/.Abs[x_]/;Simplify@Reduce[$Assumptions&&x>0]:>x

It's not particularly pretty, but it works. Thanks to ssch for the idea!

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2 Answers 2

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One workaround is to use Reduce instead, and simplify that with FullSimplify:

 assum=And @@ Thread[{a, b, c, d, e} > 0]; Reduce[assum && a + b/(c + (d - e)^2) > 0] (* Out: e > 0 && d > 0 && c > 0 && b > 0 && a > 0 *) FullSimplify[%, assum] (* Out: True *)
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  • $\begingroup$ Good idea! Should have thought of that... $\endgroup$ Commented Jul 9, 2012 at 18:10
  • $\begingroup$ I just realized that this actually doesn't help me. What I would like is Mathematica to simplify Abs[a + b/(c + (d - e)^2)] for given assumptions. $\endgroup$ Commented Jul 9, 2012 at 18:25
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Either of the below logical formulations will handle this. Generic simplifiers might or might not react well to logical expressions.

In[21]:= Resolve[ ForAll[{a, b, c, d, e}, Implies[And @@ Thread[{a, b, c, d, e} > 0], a + b/(c + (d - e)^2) > 0]]] Out[21]= True In[20]:= Resolve[ ForAll[{a, b, c, d, e}, And @@ Thread[{a, b, c, d, e} > 0], a + b/(c + (d - e)^2) > 0]] Out[20]= True
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  • $\begingroup$ That's very interesting to know! I didn't even know Resolve existed. But is there also a way to simplify Abs[...]? Of course I could do a replacement and apply your trick (or ssch's) but that's not very elegant. $\endgroup$ Commented Jul 9, 2012 at 19:24
  • $\begingroup$ Might be a straightforward way but none is coming to mind right now. $\endgroup$ Commented Jul 9, 2012 at 19:43

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