I How to start with proving ##a^a + b^b > a^b + b^a##?

[MevsEinstein]

TL;DR Summary

what the title says

When I learned about Catalan's conjecture, I morphed the expression $3^2-2^3$ to $a^b+b^a$. Then, out of curiosity, I compared $a^b+b^a$ to $a^a+b^b$. I have tried many numbers (greater that or equal to 1) for both $a$ and $b$, noticing that $a^a+b^b$ is bigger than $a^b+b^a$. I don't know how I am going to prove this since it contains unknown powers and a greater sign. What path should I consider taking to prove this inequality?

 

[Office_Shredder]

I'm pretty skeptical that you can prove Catalan's conjecture with this, but let's assume $a>b\geq 1$, and let's let $x > y \geq 0$. We will prove$a^x + b^y \geq a^y+b^x$. Your use case is $x=a$ and $y=b$.

$a^x+b^y=a^ya^{x-y}+b^y$ and $a^y+b^z= a^y+b^y b^{x-y}$. Both of these expressions are $a^y+b^y$ where we multiply one of the summands by some quantity. Since the summands are at least one and the multipliers are at least one, it is hopefully intuitive that if we multiply the larger summand by a larger number, the thing we get is larger overall. $a^y>b^y$ and $a^{x-y}>b^{x-y}$, so that completes the intuitive proof.If you want to throw more algebra at it, we need to compute exactly how much larger than $a^y+b^y$ each side is
$a^ya^{x-y}+b^y = a^y(a^{x-y}-1+1)+b^y$
$=a^y(a^{x-y}-1)+a^y +b^y$

Similarly $a^y+b^x= a^y+b^y+ b^y(b^{x-y}-1)$.

So all we have to do is prove $a^y(a^{x-y}-1) \geq b^y(b^{x-y}-1)$. Note each side is multiplying two non negative numbers, so it suffices to prove each piece of the multiplication is larger. Since $a>b$, $y\geq 0$ and $x-y\geq 0$, we get $a^y>b^y$, and $a^{x-y} \geq b^{x-y}$. Subtracting one from the last inequality completes the proof.

 

[MevsEinstein]

I'm pretty skeptical that you can prove Catalan's conjecture with this

I wasn't actually trying to prove it I was just playing around. Thanks for your proof by the way!

 

[mathwonk]

I noticed the question before noticing the nice answers, so already had this alternative answer:

a^b +b^a < a^a + b^b is equivalent to a^b - a^a < b^b - b^a. By the fundamental theorem of calculus, the left side is the area under the graph of y = e^x, between the points a.Ln(a) and b.Ln(a) an interval of length Ln(a).(b-a).

Similarly the right side is the area under the same graph between the points a.Ln(b) and b.Ln(b) an interval of length Ln(b).(b-a).

Assuming 0 < a < b, the second interval begins further to the right, and is also longer than the first interval,
so the amount of area is greater for the second interval. This holds for any increasing graph.

In this case, moving an interval to the right by an amount t>0, and keeping it the same length, increases the area by a factor of e^t > 1. and then making the interval also longer by moving the right endpoint further to the right increases the area more.