So, I have two real numbers, and I need to prove or disprove the statement: $a^b<b^a$.
Note: The question contains real numbers and not parameters. I just don't want you to solve it for me, so I don't reveal the numbers... This is a calculus I question.
I have thinking about this question for a while, with no clue on what to do.
I thought of defining some functions, and use some theorems we learned...
Any hints on how to deal with those kind of questions will be much appreciated!
Thanks a lot!
I will try to give a good answer:
First, as the question is stated, the answer is no. Because you are not ordering $a$ and $b$, so you can reverse the tuples, for example $(3,10)$ and $(10,3)$.
Even asking for $b<a$, the tuple (4,2) does not satisfy the inequality because $2^4=4^2$. You can even generate an infinite family of counterexamples with $a=t^{\frac{1}{t-1}}$ $b=t^{\frac{t}{t-1}}$.
However, as the comments above showed, with the condition $e<b<a$ then the inequality holds.
If we want to show that $a^b<b^a$, then we can raise both powers to the $\frac{1}{ab}$ power (assuming neither value is $0$) to get that our problem is equivilant to showing that
$$\left(a^{b}\right)^{\frac{1}{ab}}=a^{\frac{1}{a}}<\left(b^{a}\right)^{\frac{1}{ab}}=b^{\frac{1}{b}}$$
We now observe that
$$\lim_{x\to\infty}\ln\left(x^{\frac{1}{x}}\right)=\lim_{x\to\infty}\frac{\ln(x)}{x}=0$$
and thus
$$\lim_{x\to\infty}x^{\frac{1}{x}}=1$$
and at $x=0$ we have that $0^{\infty}=0$, and so
$$\lim_{x\to0}x^{\frac{1}{x}}=0$$
Finding extrema of $x^{\frac{1}{x}}$ yields that
\begin{align*} \frac{d}{dx}x^{\frac{1}{x}}&*=\frac{d}{dx}e^{\frac{\ln\left(x\right)}{x}}\\ &=e^{\frac{\ln\left(x\right)}{x}}\frac{d}{dx}\frac{\ln(x)}{x} &=e^{\frac{\ln\left(x\right)}{x}}\left(\frac{1}{x^{2}}-\frac{\ln\left(x\right)}{x^{2}}\right) \end{align*}
and thus our extreme point is when $\frac{d}{dx}x^{\frac{1}{x}}=0$, i.e $\ln(x)=1$, i.e $x=e$. Plugging in $x=e$ to the equation yields $e^{\frac{1}{e}}\approx 1.44>1$, and so this must be a maximum (comparing to endpoints). Thus, if both numbers $a,b$ are on the same side of $e$ then whichever is closest to $e$ has a larger value of $x^{\frac{1}{x}}$. If they are on opposite sides, then you are out of luck with this method.
As shown in the other answers, the hard case is when $a < e < b$.
Many years ago, I proved the following result in this case:
If $1 < x < e$ and $e < y < e^2/x$ (i.e., $xy < e^2$) then $y^{1/y} > x^{1/x} $.
I don't know about $y > e^2/x$.
I've reconstructed my proof and here it is.
Let $f(x) =\dfrac{x^{1/x}}{(e^2/x)^{x/e^2}} $ with $1 < x < e$.
The proof shows that $\ln(f(x)) \lt 0$ for $1 < x < e$. This reduces to showing that $h(z)=(1-z)\cosh(z)-e^{-z} \lt 0$ for $0 < z < 1$.
As usual, I'm sure there are simpler proofs than mine, which shows that the coefficients of $h(z)$ are all negative.
Show $f(x) <1 $.
$\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 $e < y < e^2/x$, $r(y) \gt r(e^2/x) \gt r(x) $.
Therefore, if $1 < x < e < y < e^2/x$, or $1 < x < e$ and $xy < e^2$ then $r(y) > r(x) $.
