I've been trying to prove $n^\frac{1}{n} > (n+1)^\frac{1}{n+1}$ and I got to an equation in the midst of it, don't know whether it's correct or not $x^\frac{1}{x} + x^{1+\frac{1}{x}} = 1/3$ for a lower boundary point for the correctness of the eq. above. I'm doing it for fun and I don't remember how to solve this type of equations. Need help! Even if the equation itself is incorrect in the context of the proof - how would I approach solving it? I can rewrite it as $3(x^2 + x) = x^{\frac{x-1}{x}}$ or $log_x{3(x^2 +x)}=\frac{x-1}{x}$ but what should I do next?

Edit: My main question is really not about the inequality, but rather about the equation I got: $$x^\frac{1}{x} + x^{1+\frac{1}{x}} = 1/3$$ is it possible to solve it by taking logs and such? Can't wrap my head around it!

Edit 2: Turns out there's a special function for this kind of equations, called Lambert W, which is an inverse of $f(z)=ze^z$! So the answer to my question is no, I cannot solve it with my high-school level math chops!

  • $\begingroup$ Let $y=x^{\frac{1}{x}}$ ... find the turning point (it should be at $x=e$) ... etc ... $\endgroup$ Commented Oct 2, 2017 at 12:01

1 Answer 1


In such symmetrical cases we generally assume a function ($y=x^{1/x}$) and try to show if it's increasing or decreasing.

For this, we'll take log on both sides and differentiate to get $$\frac{dy} {ydx}=\frac{1-lnx}{x^2}\rightarrow dy/dx=x^{1/x}\frac{1-lnx}{x}$$

It is zero at $x=e$. As observed, graph of $y$ will increase for $x<e$ and decrease for $x>e$.

Hence, $\forall n\geq3, f(n+1)<f(n)\rightarrow {n+1}^{\frac{1}{n+1}}<n^{1/n}$.

For reference: the exact graph of the function $y=x^{1/x}$:

enter image description here

  • $\begingroup$ I get it, thanks, that's a great way to prove it! But how would I solve the equation that I got for x, if it's in real numbers, like just as a separate equation? The root for it will not be $e$! It's around 0.4847... but how do I get to it? $\endgroup$ Commented Oct 2, 2017 at 12:40
  • $\begingroup$ @Non-chemist_dude We generally try to separate any two functions on the RHS and LHS of the equation, and then draw the graphs for each of them, then find intersection point. Or we might have a bounded function (eg: $sinx$) on one side. However, in this equation, $x^\frac{1}{x} + x^{1+\frac{1}{x}} = 1/3$ it's neither the case. Let me think more. $\endgroup$ Commented Oct 2, 2017 at 12:44
  • $\begingroup$ HI, Gaurang, I found the answer to my question, and it's negative - there's en.wikipedia.org/wiki/Lambert_W_function which helps solve similar equations. $\endgroup$ Commented Oct 3, 2017 at 5:44
  • $\begingroup$ @Non-chemist_dude Oh, thanks for notifying me. I haven't studied maths this deep, and am unsure of how it works. But, anyway. $\endgroup$ Commented Oct 3, 2017 at 11:38

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