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For positive integer $n$ and positive reals $x,y,a,b$

$x a^n + y b^n < (x+\frac{a}{3})\int (a+x+\frac{1+a}{b})^n da + (y+\frac{b}{3})\int (b+y+\frac{1+b}{a})^ndb+\frac{9}{4} $

Is this true ?

What is the easiest way to decide this ?

In inequalities is there a priority in proofs for calculus tools vs algebraic methods ?

I sometimes find inequalities puzzling. Some advice is welcome.

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Indefinite integrals involve an arbitrary constant of integration?? –  GEdgar Oct 24 '12 at 21:29
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1 Answer 1

  1. It is unlikely to be true since you do not give limits for the integrations. Also, you use $a$ and $b$ as both variables and variables of integration, which is probably an error.

  2. Try the asserted inequality for $n = 1$ and $2$ and see what happens.

  3. The first priority is just to prove it. The next priority is to prove it a different way. The third priority is to give a really elegant proof that shows "why" the result is really true. People seems to give extra points for a more elementary proof (i.e. geometry > algebra > calculus > cohomology theory).

Also, how did you come across this? It looks like a numerical integration formula.

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1) As an example $\int a$ $ da$ means $\frac{a^2}{2}$ , when I wrote both variable I mean the primitive with constant 0. 2) That seems not elegant. More like a computer approach. It might work for $n$ $= 1,2,3,4$ but fail at $5$. 3) I was looking for a neat trick. And I wonder if I need to derivate here. And how to begin. ..Well yes , I was thinking about numerical integration and inequalities when I realized I have no (good) tools for proving such things. –  mick Oct 24 '12 at 21:49
    
But a is also a variable outside the integral. That means that if you use da in the integral, then a inside the integral can be either the a associated with da or the a from outside the integral. –  marty cohen Oct 27 '12 at 6:11
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