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Thermodynamic Inequality

  1. Aug 10, 2013 #1
    Prove that
    $$ \frac{y^x-1}{xy^{x-1}(y-1)}<1$$
    where [itex]x,y \in ℝ, x>1 [/itex] and [itex]y>1.[/itex]

    I was able to prove it using calculus, but am wondering if there was another way of doing so, like exploiting some inequality-theorems which involves real numbers. I'll be glad if anyone can show me a way and quench my hunch.

    Cheers.
     
  2. jcsd
  3. Aug 10, 2013 #2

    Simon Bridge

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    $$ \frac{y^x-1}{xy^{x-1}(y-1)}<1$$
    where [itex]x,y \in ℝ, x>1 [/itex] and [itex]y>1.[/itex]

    hmmm, not a fancy theorem but ... did you try - subtract 1 from both sides and put the LHS over a common denominator:

    which means that either the numerator or the denominator is negative.

    notice that the denominator is a product of only positive numbers?
    see where this is going?
     
  4. Aug 11, 2013 #3
    Thanks. But I am afraid I still can't see how the numerator is negative; that is:
    $$y^x-1-xy^x+xy^{x-1}<0$$.
    I was stuck with this expression before you pointed out. I think this part is the difficult one.
     
  5. Aug 11, 2013 #4

    Simon Bridge

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    It is sometimes easier to see when something is negative ... you may have to experiment a bit to find a pattern... i.e. if x=1, what values of y will make the relation = 0, what values >0 and what values <0?

    Alternatively:
    $$\Rightarrow y^x-1 < xy^{x-1}(y-1)$$
    ... what conditions have to be satisfied for that to be true?
    You should be able to see how it will be true just by graphing the LHS and the RHS.
    Try figuring what y and x could be - like what happens is x>1 and 0<y<1 ?

    geometrically, the LHS and RHS describe surfaces in 3D - LHS=RHS will be the intersection of these two surfaces.
     
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