Prove $(x-1)(y-1)>(e-1)^2$ where $x^y=y^x$, $y>x>0$.

Question

Let $x,y$ be different positive real numbers satisfying $x^y=y^x$. Prove $(x-1)(y-1)>(e-1)^2$.

We may suppose $a=\frac{y}{x}>1$. Then we obtain $x=a^{\frac{1}{a-1}},y=a^{\frac{a}{a-1}}.$ But how to go on?

Answer 1

If you know that $$x^y=y^x \implies x=-\frac{y}{\log (y)}W\left(-\frac{\log (y)}{y}\right)$$ where $W(t)$ is Lambert function, then $$(x-1)(y-1)=(1-y)\Bigg[1+ \frac{y}{\log (y)}W\left(-\frac{\log (y)}{y}\right)\Bigg]=f(y)$$

$f'(y)$ cancels at $y=e$ and, at this point, $x=e$. The second derivative $$f''(e)=\frac{2 (2 e-5)}{3 e}\sim 0.107069~~ >~~ 0$$ confirms that this point is a minimum and $f(e)=(e-1)^2$.

Edit

As @Martin R commented, I did not prove that this is the global minimum.

In fact, this can be done using the fact that $\forall t \geq e$ $$W(t) \geq\log (t)-\log (\log (t))+\frac 12\frac{\log (\log (t))}{ \log (t)}=f(t)$$ $$W(t) \leq \log (t)-\log (\log (t))+\frac e{e-1}\frac{\log (\log (t))}{ \log (t)}=g(t)$$ or even better tha approximation of $W(t)$ I gave here.

Answer 2

Continuing your idea, let $a = 1+ \dfrac 1t, t > 0.$ Then, you can write your inequality as $xy-x-y > e^2-2e$, which is in terms of the variable $t:$

$$f(t) = t\ln\left(1+\frac 1t\right) + \ln\left(\left(1+\frac 1t\right)^{t+1}-2-\frac 1t\right) > 1+\ln(e-2).$$

Wolfram says that this is indeed a decreasing function in $t$, so the infimum will be: $$\lim\limits_{t\to\infty} f(t) = 1+ \ln(e-2).$$

Then, the only thing to show now is the fact that $f(t)$ is decreasing, but I suspect this will take an ugly computation.

EDIT: Actually, the derivative did not turn out to be too bad:

\begin{align}f'(t) &= \\ &\dfrac{\dfrac{1}{t^2} + \left(1+\dfrac 1t\right)^{t+1}{\left(2\ln\left(1+\frac 1t\right)-\dfrac 1t-\dfrac{1}{t+1}\right)} + \left(2+\dfrac{1}{t}\right) \left(\dfrac{1}{1+t} - \ln\left(1+\dfrac{1}{t}\right)\right)}{\left(1+\dfrac 1t\right)^{t+1} - \dfrac 1t-2} & \\<0\end{align}

Finally, the above is equivalent to:

$$\left(1+\dfrac 1t\right)^{t+1} > \dfrac{\dfrac{1}{t^2} + \left(2+\dfrac 1t\right)\left(\dfrac{1}{1+t} - \ln\left(1+\dfrac{1}{t}\right)\right)}{\dfrac{1}{t}+\dfrac{1}{1+t} - 2\ln\left(1+\dfrac{1}{t}\right)},$$

which I think should be doable by brute-force expanding with a Taylor's series etc.

Answer 3

(A comment rather than an answer but much easier to enter)

This is an improvement of my answer Proving or disproving: $a^b<b^a$? where I showed that if $x^{1/x} = y^{1/y}$ and $1 < x < y$ then $y > e^2/x$.

This is an improvement since $(e-1)^2/(x-1)+1- e^2/x =(e - x)^2/((x - 1) x) \gt 0$.

Answer 4

Sketch of a proof:

Fact 1: Let $f(u) = \frac{\ln u}{u}$. Then, $f(u)$ is strictly increasing on $(0, \mathrm{e})$, and strictly decreasing on $(\mathrm{e}, \infty)$.

WLOG, assume $y < x$. From $x^y = y^x$, we have $\frac{\ln x}{x} = \frac{\ln y}{y}$. By Fact 1, it is easy to prove that $1 < y < \mathrm{e} < x$.

We need to prove that $y > 1 + \frac{(\mathrm{e} - 1)^2}{x-1} \triangleq y_0$. Clearly, $y_0\in (1, \mathrm{e})$. By Fact 1, it suffices to prove that $f(y) > f(y_0)$ or $\frac{\ln x}{x} > f(y_0)$. It suffices to prove that $$\frac{\ln x}{x} > \frac{\ln\left( 1 + \frac{(\mathrm{e} - 1)^2}{x-1}\right)}{1 + \frac{(\mathrm{e} - 1)^2}{x-1}}, \ \forall x > \mathrm{e}.$$ It is true. But my proof is not nice. (Hint: Use Fact 2, then use derivative.)

Fact 2: $\frac{\ln v}{v} \ge \frac{(2\mathrm{e} + 10)v - 16\mathrm{e} + 4\mathrm{e}^2} {(3\mathrm{e} - 3)v^2 + (16\mathrm{e} - 4\mathrm{e}^2)v - 19\mathrm{e}^2 + 7\mathrm{e}^3}$ for all $v \ge \mathrm{e}$. (Hint: Take derivative.)

Remark 1: I propose the following problem:

Let $0 < y < x$ such that $\frac{\ln x}{x} = \frac{\ln y}{y} = a$ ($a>0$). Prove that $$(x-1)(y-1) > \left(\mathrm{e} - 1 + \frac{5}{8}\left(\frac{1}{\mathrm{e}} - a\right)\right)^2.$$

By the way, there are many similar problems in MSE or AoPS. These problems has the following description: Let $f(z)$ be a unimodal function. Let $f(x) = f(y) = a$. Then $g(x, y) \ge h(a)$.

See my closed post: Inequalities involving roots of some functions (e.g., $\frac{\ln x}{x}$, $x\ln x$)
For example, Estimate the bound of the sum of the roots of $1/x+\ln x=a$ where $a>1$,
let $f(x) = (x-1)\ln x$, and given $0 < a < b$. If $f(a) = f(b)$, prove that $\frac{1}{\ln a}+\frac{1}{\ln b} < \frac{1}{2}$