Is There a Non-Calculus Method to Prove this Thermodynamic Inequality?

[emroz92]

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.

 

[Simon Bridge]

$$ \frac{y^x-1}{xy^{x-1}(y-1)}<1$$
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?

 

[emroz92]

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.

 

[Simon Bridge]

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.

 

[CellCoree]

I understand the importance of finding multiple ways to prove a hypothesis or inequality. In this case, we can use the inequality theorem of real numbers to prove the thermodynamic inequality given above.

First, let's rearrange the inequality to make it easier to work with:
$$ y^x < xy^{x-1}(y-1) + 1 $$

Now, we can use the inequality theorem which states that for any real numbers a and b, if a < b, then for any positive number c, ac < bc.

In this case, we can let a = y and b = x, and then apply the inequality theorem to get:
$$ y^x < xy^{x-1} $$

Next, we need to show that $xy^{x-1} < xy^{x-1}(y-1) + 1$. This can be done by using the inequality theorem again, but this time with a = 1 and b = y-1. This gives us:
$$ y^{x-1} < y^{x-1}(y-1) + 1 $$

Substituting this into our original inequality, we get:
$$ y^x < xy^{x-1} < xy^{x-1}(y-1) + 1 $$

Therefore, we have proven that the thermodynamic inequality holds for all x,y \in ℝ, x>1, and y>1.

In conclusion, we can use the inequality theorem of real numbers to prove the thermodynamic inequality given in the prompt. This approach may be more straightforward and does not require the use of calculus.