Prove that $e^x > x^x y^y (y + 1)$, where $x$ and $y$ are positive real numbers satisfying $x^x + y^y + y - x = 2$

Question

Problem$\,$ Let $x$ and $y$ be positive real numbers such that $x^x+y^y+y-x=2.$ Prove that: $$e^x>x^xy^y(y+1)$$

This problem is about the application of the popular inequality $e^x\ge x+1.$ So, we can not use calculus or derivative.

Since $x$ is positive, so the inequality $e^x\ge x+1$ is strict. It suffices to prove that $$x+1>x^xy^y(y+1)$$

My difficulty is that, I cannot express the term $x$ in terms of $y$ using $$x^x+y^y+y-x=2.$$

It is the point I stuck. Because, the given condition is not a polynomial.

But $e^x\ge x+1\ge x^xy^y(y+1)$ will not give strict inequality. So it doesn't suffice. — Dietrich Burde, Jun 04 '25 at 18:06
@DietrichBurde Yes. I thought that I can analyse the equality case later. — user559619, Jun 04 '25 at 18:11
You also can do it now, not later, see here. For positive $x$ we do have $e^x>1+x$. — Dietrich Burde, Jun 04 '25 at 18:12
@DietrichBurde I think that no. $x=y=1$ is indeed trivial and works $e>2$. But we can think about $x\to 0^+$ and $y\to 0^+$, which is not trvial. — user559619, Jun 04 '25 at 18:23
Note that $x = x^x + y^y + y - 2 = (x^x-1) + (y^y-1) + y$. Plug that into $e^x$. — Dermot Craddock, Jun 04 '25 at 19:07

score 5 · Accepted Answer · answered Jun 04 '25 at 19:43

Isolating $x$ in the condition $x^x + y^y + y - x = 2$ yields \begin{equation} x = x^x + y^y + y - 2 = (x^x - 1) + (y^y - 1) + y\,. \end{equation} Therefore \begin{equation} e^x = \underbrace{e^{x^x-1}}_{\geqslant x^x} \cdot \underbrace{e^{y^y-1}}_{\geqslant y^y} \cdot \underbrace{e^y}_{> y+1} > x^xy^y(y+1) \end{equation} as claimed.

score 2 · Answer 2 · answered Jun 05 '25 at 01:29

Alternative proof.

By AM-GM (twice), we have \begin{align*} x^xy^y (y + 1) &\le \frac{(x^x + y^y)^2}{4}(y + 1)\\ &= \frac{(2 + x - y)^2}{4}(y + 1)\\ &= \frac{2 + x - y}{2} \cdot \frac{2+x-y}{2} \cdot (y+1)\\ &\le \left(\frac{\frac{2 + x - y}{2} + \frac{2+x-y}{2} + (y+1)}{3}\right)^3\\ &= (1+x/3)^3. \end{align*}

It suffices to prove that $\mathrm{e}^x > (1 + x/3)^3$, or $\mathrm{e}^{x/3} > 1 + x/3$ which is true. We are done.

Prove that $e^x > x^x y^y (y + 1)$, where $x$ and $y$ are positive real numbers satisfying $x^x + y^y + y - x = 2$

2 Answers2