TITLE: help with a simple equation with powers and logs
QUESTION [3 upvotes]: 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!

REPLY [4 votes]: 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 $y=x^{1/x}$:
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