# $\forall x,y>0, x^x+y^y \geq x^y + y^x$

Prove that $\forall x,y>0, x^x+y^y \geq x^y + y^x$

A friend of mine told me none of the teachers in my school have succeeded in proving this seemingly simple inequality (it was asked at an oral exam last year). I tried it myself, but I made no significant progress.

• I am assuming you are talking about $x,y\in\mathbb{R}$, correct? – Daniel W. Farlow May 28 '15 at 16:09
• yes but it must be $x,y>0$ – Dr. Sonnhard Graubner May 28 '15 at 16:14
• @Dr.SonnhardGraubner How do you know? You're not the OP. If $x,y\in\mathbb{Z}$, then perhaps a double induction solution may be apropos, but I suspect $x,y\in\mathbb{R}$. OP needs to clarify. – Daniel W. Farlow May 28 '15 at 16:20
• yes it is true, but i know the problem – Dr. Sonnhard Graubner May 28 '15 at 16:21
• @LeGrandDODOM Perhaps you may add some 'english' in your title ? Because as stated is difficult to find your question in 'search'. – Kenobi May 29 '15 at 19:41

## 2 Answers

WLOG, let $x \ge y$.

Case 1: Let $x \ge 1$, then for $t > 0$, $x^t \log x \ge y^t \log y$ $$\implies \int_y^x \left(x^t \log x - y^t \log y \right)dt \ge 0 \implies (x^x - x^y) - (y^x-y^y) \ge 0$$

Added
Case 2: For $x< 1$ - let $r = \dfrac{x}y\ge 1$. Then the inequality is: $$\frac{y^x}{x^y+y^x}r^x+\frac{x^y}{x^y+y^x}\frac1{r^y} \ge 1$$

Using weighted AM-GM, this reduces to showing: $$xy^x \ge yx^y \iff \frac{\log x}{1-x} \ge \frac{\log y}{1-y}$$ which is obvious.

Since the statement is symmetric in $x$ and $y$, it suffices to show: for fixed $y>0$, the quantity $x^x+y^y-x^y-y^x$ is nonnegative for $x\ge y$. And since this quantity vanishes at $x=y$, it suffices to show that it is increasing for $x\ge y$. But its derivative (with respect to $x$) is $$x^x \log x + x^x - yx^{y-1} - y^x\log y = (x^x \log x - y^x \log y) + (x^x - yx^{y-1});$$ it now suffices to show that both quantities in parentheses are positive for $x\ge y>0$.

But for fixed $x>0$, the functions $t^x\log t$ and $tx^{t-1}$ are both increasing functions of the positive variable $t$: the derivative of the first function is $t^{x-1}+t^x\log^2t$ which is clearly positive; while the derivative of the second function is $x^{t-1}+tx^{t-1}\log t$, which is positive because $t\log t \ge -e^{-1} > -1$ for all $t$. In particular, their values at $t=x$ exceed their values at $t=y$, which confirms that both quantities are positive.

EDIT: Somehow I got both these last derivatives wrong, as LeGrandDODOM pointed out. The actual derivatives of $t^x\log t$ and $tx^{t-1}$ are, respectively, $t^{x-1}(x\log t+1)$ and $x^{t-1}(t\log x+1)$, neither of which is forced to be positive. I'm making this answer community wiki in case others can see how to rescue the idea.

• It seems the derivative of $t\to t^x \log(t)$ is $t^{x-1}(x\log(t)+1)$, which needs not be positive ? – Gabriel Romon May 28 '15 at 16:47
• And the derivative of $t\to tx^{t-1}$ seems to be $x^{t-1}+tx^{t-1}\log x$ – Gabriel Romon May 28 '15 at 16:51