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Let $0<x<y$, such that $$x^y=y^x$$ show that $$x+y>2e$$

Since $$y\ln{x}=x\ln{y}\Longrightarrow \dfrac{\ln{y}}{y}=\dfrac{\ln{x}}{x}$$ Let $$f(x)=\dfrac{\ln{x}}{x}\Longrightarrow f'(x)=\dfrac{1-\ln{x}}{x^2}$$ If $0<x<e$ then $f'(x)>0$, if $x>e$,then $f'(x)<0$

$\Longleftrightarrow y-e+x-e>0$,maybe consider $f(x-e)$ and $f(y-e)$,wihch bigger? I just do it now

Thanks in advance!

partofsha
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  • you could try studying the function $g(t)= f(e+t)-f(e-t)$, if you show that $g(t)<0$ then you are done. I think you can do that, but I have not made the calculation myself. So the idea is try show $g'(t)<0$. – clark May 04 '16 at 14:46
  • But $x+e$ and $e-x$ maybe no in same Increasing (or creaing) range – partofsha May 04 '16 at 14:58

3 Answers3

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We can see that $$f(x)=\frac{\ln x}{x}$$ is strictly increasing in $(0,e)$ and striclty decreasing in $(e,+\infty)$.

Thus the equality $f(x)=f(y)$ cannot hold for the case of $x,y \in (0,e)$ or $x,y \in (0,+\infty)$.

Since $x\lt y$, it follows that $x\in (0,e)$ and $y \in (e,+\infty)$.

So let $y=rx$, for $r\in (1,+\infty)$ and we obtain $$(xr)^x = (x)^{rx}\implies rx \ln x = x (\ln r + \ln x)\implies(r−1)\ln x = \ln r$$

So, $$x = r^{1/(r−1)}, y= r^{r/(r−1)}$$ and we need to show that for $r\in (1,+\infty)$, the function $$g(r)=r^{1/(r−1)}+r^{r/(r−1)}\gt2e$$

We notice that $$\lim_{r\rightarrow1}r^{1/(r−1)}=\lim_{r\rightarrow1}r^{r/(r−1)}=e$$

Indeed, if we set $u=\frac1r$ we get $\lim(1+\frac1u)^u=e$, and similarly for the second limit.

Thus we only need to show that $g(r)$ is strictly increasing in $(1,+\infty)$.

EDIT
(As clark rightly pointed, the first approach to prove that $g(r)$ is strictly increasing was wrong on my part-see the comments below. I hope this new take is correct.)


We wish to show that $g(r)=r^{\frac{1}{r-1}}(r+1)$ is strictly increasing.

We will show that $h(r)=\ln g(r)$ is strictly increasing and since $\ln x$ is s.i, we derive from this the fact that $g(r)$ is s.i. (That is, if $h(x)=\ln g(x)$ is strictly increasing $\Rightarrow g(x)$ is strictly increasing. To show this one can simply use Reductio ad absurdum and assume $h(x)$ is s.i while $g(x)$ is not s.i. )

So, we have $$\ln g(r)=\ln r^{\frac{1}{r-1}}(r+1)=\ln r^{\frac{1}{r-1}}+\ln(r+1)=\frac1{r-1}\ln r+\ln (r+1)\\h'(r)=\ln r+\frac{1}{r(r-1)}+\frac1{r+1}\gt0, \forall r \in (1,+\infty)$$ It follows that $h(r)$ is strictly increasing and so is $g(r)$.

MathematicianByMistake
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    The derivative $g'(r)\not=x$ since $x$ depends on $r$. – clark May 04 '16 at 15:36
  • @clark Does it depend on it? $y$ certainly does, but in respect to $x$, $r$ is just a parameter. Am I missing something here? If so I will edit the answer and hope that $g(r)$ can be shown to be strictly positive otherwise. in any case, thank you for the heads-up! – MathematicianByMistake May 04 '16 at 15:51
  • since $g'(r)\neq x$,so maybe $g'(r)>0$ is not easy to prove it? – partofsha May 04 '16 at 15:53
  • @partofsha If we can show that $g'(r)\gt0$ we are done, but I am having doubts because of clark' s observation. We can try tackling it straightforwardly but it seems very dificult. – MathematicianByMistake May 04 '16 at 15:56
  • We want to show that the function $g(r)=r^{1/(r−1)}+r^{r/(r−1)}$ is increasing. If you differentiate in respect to $r$ you will not get $g'(r)=x$(substitue back what $x=r^{\frac{1}{r-1}}$), since that would be equal to $g'(r)=x+x'(r+1)$ by the product rule – clark May 04 '16 at 15:57
  • @clark Yes, I see it now. Thanks again! I will edit the answer. – MathematicianByMistake May 04 '16 at 15:58
  • I found $x=r-1$,then $g(r)=g(x)=(1+x)^{1+\frac{1}{x}}(x+1)$,then we have $$f'(x)=(1+x)^{\frac{1}{x}-1}\cdot\dfrac{x(x^2+2x+2)-(x^2+3x+2)\ln{(x+1)}}{x^2} $$,so we only prove $x(x^2+2x+2)-(x^2+3x+2)\ln{(x+1)}>0,x>0$，maybe at last inequality is not hard to prove it – partofsha May 04 '16 at 16:07
  • because $$ (x(x^2+2x+2)-(x^2+3x+2)\ln{(x+1)})'=3x(x+1)-(3x+2)\ln{(x+1)}>3x(x+1)-(3x+2)\cdot x=x>0$$,where use $\ln{(x+1)}<x$ – partofsha May 04 '16 at 16:12
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As MathematicianByMistake answer,we want prove $$g(r)=r^{\frac{1}{r-1}}+r^{\frac{r}{r-1}}=(x+1)^{\frac{1}{x}}+(x+1)^{\frac{x+1}{x}}=(x+1)^{\frac{1}{x}}(x+2),x>0$$ since $$g'(x)=(x+1)^{\frac{1}{x}-1}\cdot\dfrac{x(x^2+2x+2)-(x^2+3x+2)\ln{(x+1)}}{x^2}$$ we only prove $$h(x)=x(x^2+2x+2)-(x^2+3x+2)\ln{(x+1)}>0,x>0$$ since $$h'(x)=3x(x+1)-(3x+2)\ln{(x+1)}>3x(x+1)-x(3x+2)=x>0$$

partofsha
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Alternative solution:

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

By Fact 1, it is easy to prove that $1 < x < \mathrm{e} < y$.

We need to prove that $y > 2\mathrm{e} - x$.

Since $2\mathrm{e} - x \in (\mathrm{e}, \infty)$, by Fact 1, it suffices to prove that $f(y) < f(2\mathrm{e} - x)$.

Since $f(x) = f(y)$, it suffices to prove that $f(x) < f(2\mathrm{e} - x)$ that is $$\frac{\ln x}{x} < \frac{\ln (2\mathrm{e} - x)}{2\mathrm{e} - x}.$$ It suffices to prove that, for all $x$ in $(1, \mathrm{e})$, $$x\ln (2\mathrm{e} - x) - (2\mathrm{e} - x)\ln x > 0. $$ Denote LHS by $g(x)$. We have $$g'(x) = \ln (2\mathrm{e} - x) - \frac{x}{2\mathrm{e} - x} + \ln x - \frac{2\mathrm{e} - x}{x}$$ and $$g''(x) = \frac{2(\mathrm{e} - x)(4\mathrm{e}^2 + 2\mathrm{e} x - x^2)}{(2\mathrm{e} - x)^2x^2}.$$ Since $g''(x) > 0$ for all $x$ in $(1, \mathrm{e})$ and $g'(\mathrm{e}) = 0$, we have $g'(x) < 0$ for all $x$ in $(1, \mathrm{e})$.

Since $g(\mathrm{e}) = 0$, we have $g(x) > 0$ for all $x$ in $(1, \mathrm{e})$.

We are done.

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

River Li
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