3

Let $x,y>0$, it seems (with numerical simulations) that $x^x+y^y \geq x^y +y^x$.

If this is true, it has to be well known by some people.

Does this inequality have a name? several proofs?

Martin Sleziak
  • 53,687
user254351
  • 31
  • It's like rearrangement inequality, but with powers. But I can't see how to show it... – ajotatxe Jul 13 '15 at 22:08
  • Check http://math.stackexchange.com/questions/1302691 – Macavity Jul 14 '15 at 01:23
  • More http://math.stackexchange.com/questions/482551/an-exponential-rearrangement-inequality-xxyyxyyx?s=4|6.7900 and http://math.stackexchange.com/questions/410895/inequality-with-exponents – b yen Jul 14 '15 at 02:01

1 Answers1

1

Fix $y > 0$, and let $$f(x) = x^x - x^y - y^x.$$ We are done if we can show the minimum of the $f$ is $-y^y$. Computing the derivative, $$f'(x) = -y x^{y-1}+x^x (\log (x)+1)-y^x \log (y).$$ $f'$ has a zero when $x = y$ and changes from positive to negative there. Thus $f$ has a local minimum. But $f$ increases for $x > y$, so $f(y) = -y^y$ is the global minimum.

user217285
  • 5,735