If $1 < x < e < y$ and $x^{1/x} = y^{1/y}$ then $y > e^2/x$.

Question

If $1 < x < e < y$ and $x^{1/x} = y^{1/y}$ then $y > e^2/x$.

Since $x^{1/x}$ is decreasing for $x > e$, this is equivalent to $x^{1/x} \lt (e^2/x)^{x/e^2} $for $1 < x < e$.

Also, it would be nice to find a closer approximation to $y$ than $e^2/x$.

Also $^2$, I realize that this might be better solved using the Lambert-W function.

So, if you want to make a question of this, it could either be "Is there a simpler prrof of this result" or "Is there a better bound for the $y > e$ with $x^{1/x}=y^{1/y}$ than $e^2/x$"?

$f(x) =\dfrac{x^{1/x}}{(e^2/x)^{x/e^2}} $ with $1 < x < e$.

Want to show $f(x) <1 $.

This proof is much more involved than I like.

$\begin{array}\\ g(x) &=\ln(f(x))\\ &=(1/x)\ln(x)-(x/e^2)\ln(e^2/x)\\ &=(1/x)\ln(x)-(x/e^2)(\ln(e^2)-\ln(x))\\ &=(1/x)\ln(x)-(2x/e^2)+(x/e^2)\ln(x)\\ &=(1/x+x/e^2)\ln(x)-(2x/e^2)\\ &=(e^{-y}+e^y/e^2)y-(2e^y/e^2) \qquad x = e^y, 0 < y < 1\\ &=(e^{-y}+e^{y-2})y-2e^{y-2}\\ &=e^{-1}(e^{-y+1}+e^{y-1})y-2e^{y-2}\\ &=\dfrac{2y}{e}\dfrac{e^{-y+1}+e^{y-1}}{2}-2e^{y-2}\\ &=\dfrac{2y}{e}\cosh(-y+1)-\dfrac{2}{e}e^{y-1}\\ &=\dfrac{2}{e}(y\cosh(-y+1)-e^{y-1})\\ &=\dfrac{2}{e}((1-z)\cosh(z)-e^{-z}) \qquad z = 1-y, y = 1-z, 0 < z < 1\\ &=\dfrac{2}{e}h(z) \qquad h(z)=(1-z)\cosh(z)-e^{-z}\\ &=\dfrac{2}{e}\left((1-z)\sum_{n=0}^{\infty} \dfrac{z^{2n}}{(2n)!}-\sum_{n=0}^{\infty} \dfrac{(-1)^nz^n}{n!}\right)\\ &=\dfrac{2}{e}\left(\sum_{n=0}^{\infty} \dfrac{z^{2n}}{(2n)!}-z\sum_{n=0}^{\infty} \dfrac{z^{2n}}{(2n)!}-\sum_{n=0}^{\infty} \dfrac{z^{2n}}{(2n)!}+\sum_{n=0}^{\infty} \dfrac{z^{2n+1}}{(2n+1)!}\right)\\ &=\dfrac{2}{e}\left(-\sum_{n=0}^{\infty} \dfrac{z^{2n+1}}{(2n)!}+\sum_{n=0}^{\infty} \dfrac{z^{2n+1}}{(2n+1)!}\right)\\ &=\dfrac{2}{e}\left(\sum_{n=0}^{\infty} \dfrac{z^{2n+1}(1-(2n+1))}{(2n+1)!}\right)\\ &=-\dfrac{2}{e}\sum_{n=0}^{\infty} \dfrac{2nz^{2n+1}}{(2n+1)!}\\ &\lt 0 \qquad\text{for } z > 0\\ \end{array} $

Therefore $g(x) < 0$ so $f(x) < 1$.

Therefore, if $r(x) = x^{1/x}$, then, if $1 < x < e$, $r(x) \lt r(e^2/x) $.

Since $e^2/x > e$, if $e < y \le e^2/x$, $r(y) \ge r(e^2/x) \gt r(x) $.

Therefore, if $1 < x < e < y \le e^2/x$, or $1 < x < e$ and $xy \le e^2$ then $r(y) > r(x) $.

Therefore if $r(y) = r(x)$ then $y > e^2/x$.

Answer 1

Let $a := \frac{y}{x} > 1$.

From $x^{1/x} = y^{1/y}$, we have $x^{y/x} = y$ and $x^a = ax$ and $x^{a-1} = a$ which results in $$x = a^{\frac{1}{a-1}}. \tag{1}$$ Then, we have $$y = a x = a^{\frac{a}{a-1}}. \tag{2}$$

From (1) and (2), we have $$xy = a^{\frac{1}{a-1}} \cdot a^{\frac{a}{a-1}} = a^{\frac{a+1}{a-1}}.\tag{3}$$

It suffices to prove that $$a^{\frac{a+1}{a-1}} > \mathrm{e}^2$$ or $$F(a) := \ln a - \frac{2(a-1)}{a+1} > 0.$$

We have $$F'(a) = \frac{(a-1)^2}{a(a+1)^2}.$$ Thus, $F'(1) = 0$, and $F'(a) > 0$ for all $a > 1$. Also, $F(1) = 0$. Thus, we have $F(a) > 0$ for all $a > 1$.

We are done.

Answer 2

Writing $s=\log x$ and $t=\log y$, and defining $f(u) = u/e^u$, allows us to change this multiplicative statement into an additive statement: If $0<s<1<t$ and $f(s)=f(t)$, then $s+t>2$. This will allow us to more easily use convexity-type arguments.

We may certainly assume that $t\le2$ or else the conclusion is trivially true. To prove the statement, write $$ 0 = f(t) - f(s) = \int_s^t f'(u)\,du = \int_s^{2-t} f'(u)\,du + \int_{2-t}^1 f'(u)\,du + \int_1^t f'(u)\,du. $$ (Note that the first integral is well defined no matter what under the usual convention $\int_s^{2-t} f'(u)\,du = -\int_{2-t}^s f'(u)\,du$.) Equivalently, \begin{align*} \int_{2-t}^s f'(u)\,du &= \int_{2-t}^1 f'(u)\,du + \int_1^t f'(u)\,du \\ &= - \int_t^1 f'(2-v)\,dv + \int_1^t f'(u)\,du = \int_1^t \bigl( f'(v) + f'(2-v) \bigr) \,dv. \tag{$*$} \end{align*} Now note that $f'''(u) = (3-u)e^{-u}$, which is positive for $0\le u\le 2$. Therefore $f'(u)$ is strictly convex on $[0,2]$. In particular, for all $v\in(1,2)$, $$ \frac{f'(v) + f'(2-v)}2 > f'\biggl( \frac{v+(2-v)}2 \biggr) = f'(1) = 0. $$ Therefore the integrand on the right-hand side of ($*$) is always positive, and the endpoints of integration are in the "right order" ($t>1$); it follows that the integral is itself positive, so that ($*$) implies $$ \int_{2-t}^s f'(u)\,du > 0. $$ On the other hand, $f'(u) = (1-u)e^{-u}$ is always positive for $0<u<1$, and so we conclude that the endpoints of this integral are also in the "right order", meaning that $2-t<s$—which is the inequality $s+t>2$ we want to prove.

Answer 3

I came up with this much simpler proof.

$\begin{array}\\ g(x) &=\ln(f(x))\\ &=(1/x)\ln(x)-(x/e^2)\ln(e^2/x)\\ &=(1/x)\ln(x)-(x/e^2)(\ln(e^2)-\ln(x))\\ &=(1/x)\ln(x)-(2x/e^2)+(x/e^2)\ln(x)\\ \\ g(e) &=(1/e)\ln(e)-(2e/e^2)+(e/e^2)\ln(e)\\ &=0\\ \\ g'(x) &=((1/x)\ln(x)-(2x/e^2)+(x/e^2)\ln(x))'\\ &=\dfrac{(x^2-e^2) (\ln(x) - 1)}{e^2 x^2} \qquad\text{(according to Wolfram Alpha)}\\ &>0 \qquad \text{for } 1 < x < e\\ \\ g'(e) &=0\\ \end{array} $

Therefore $g(x) < g(e) =0$ for $1 < x < e$.