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Here's my try.

The inequality above is equivalent to $$|m^{\frac{1}{n}} - n^{\frac{1}{m}}|> \frac{1}{mn}$$ First, I want to get rid of the absolute value.

Assume without loss of generality that $m>n$. Then $m^m > n^n$. Raising this inequality to the power of $\frac{1}{mn}$ gives me $m^{\frac{1}{n}} > n^{\frac{1}{m}}$.

Next, I want to arrange everything so that it looks like $n^{\frac1n}-m^{\frac1m}$($m$ goes with $m$, $n$ goes with $n$ $-$ that's easier to consider). For that, I know when the function $f(x) = x^{\frac1x}$ is decreasing. As $f'(x) = (e^{\frac1x \cdot \ln{x}})' = x^{\frac1x - 2}\cdot (1- \ln{x})$, it is clear that $f(x)$ is decreasing on the ray $[e;+\infty )$. So, consider $m$, $n$ $\geq 3$. (Other cases can be checked numerically.)

So, $m>n$, then $m^{\frac{1}{n}} > n^{\frac{1}{n}}$. At the same time $n^{\frac{1}{m}} < m^{\frac{1}{m}}$ $\Rightarrow$ $-n^{\frac{1}{m}} > -m^{\frac{1}{m}}$. Adding up these two inequalities gives me $$m^{\frac{1}{n}} - n^{\frac{1}{m}} > n^{\frac{1}{n}} - m^{\frac{1}{m}} > 0$$ Secondly, I want to utilize the fact that $m$ and $n$ are natural. As $m>n$, then $m-n\geq 1$. Dividing by $mn$ gives me $$\frac{m-n}{mn} = \frac1n - \frac1m \geq \frac1{mn}$$ Now I just have to prove this: $$n^{\frac{1}{n}} - m^{\frac{1}{m}} > \frac1n - \frac1m$$ Or, equivalently, $$ n^{\frac{1}{n}} - \frac1n > m^{\frac{1}{m}} - \frac1m$$ So, consider the function $g(x) = x^{\frac1x} - \frac1x $. I want to show that it is decreasing, starting at some point. Taking the derivative does nothing good: $g'(x) = \frac1{x^2} \cdot (x^{\frac1x} - \ln{x} \cdot x^{\frac1x} + 1)$, and I am unable to find the zeroes of it. WolframAlpha says that $g(x)$ is, indeed, decreasing from $x\approx 5.677$, and that is good, we can always check the answer for $m$, $n$ $\leq 5$. However, this is not a satisfactory solution.

element
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1 Answers1

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Suppose that $m>n$, as a result of which $m^m>n^n$. If $m=2$, then $n=1$ and the proof is easy to complete; suppose thus that $m\ge 3$.

Applying Lagrange's mean value theorem to the function $f(x)=x^{1/(mn)}$ on the interval $[n^n,m^m]$, we get $$ \sqrt[n]m-\sqrt[m]n = f(m^m)-f(n^n) = \frac1{mn}\,c^{-1+1/(mn)}(m^m-n^n), $$ where $c\in(n^n,m^m)$. Consequently, $$ \sqrt[n]m-\sqrt[m]n > \frac1{mn}\,m^{-m+1/n}\, (m^m-n^n), $$ and to complete the proof it suffices to show that $$ m^m-n^n > m^{m-1/n}. $$ Clearly, this will follow from $$ m^m-n^n>m^{m-1/m}, $$ and then, in view of $m^{1/m}>1+1/m$ (which is very easy to prove), from $$ \left(1+\frac1m\right)(m^m-n^n)>m^m. $$ The last inequality simplifies to $$ m^{m-1} > \left(1+\frac1m\right)n^n, $$ and this is true in view of $$ m^{m-1}\ge (n+1)^n \ge \left(1+\frac1n\right)n^n > \left(1+\frac1m\right)n^n. $$

W-t-P
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  • How to prove $m^{\frac1m} > 1 + \frac1m $? I don't find it easy. – element Mar 21 '19 at 13:10
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    @Max: this is equivalent to $(1+1/m)^m<m$. The sequence $(1+1/m)^m$ is monotonically increasing and converges to Euler's number $e\approx 2.71828$. Thus $(1+1/m)^m<e<m$. – W-t-P Mar 21 '19 at 19:33
  • @Max: equivalently, use the standard and well-known inequality $e^x>1+x\ (x>0)$ to get $m^{1/m}>e^{1/m}>1+1/m$. – W-t-P Mar 21 '19 at 19:40
  • @W-t-P or binomial theorem: ${m \choose k}\frac{1}{m^k} \le 1$ for each $k$. – mathworker21 Mar 28 '19 at 21:18