Help me proving $x^{\frac{1}{x}} \geq \frac{1}{3}$ How can I show that $x^{\frac{1}{x}}\ge1/3$ is satisfied for all $x\ge b$ where $b>1/2$. One way of doing this is showing the derivative of $x^{\frac{1}{x}}$ is positive for $x>b$; however I have some difficulties proving that such $b$ must exists. Hints?
 A: Let $f(x)=x^{\frac 1x}\ \ (x\gt 0)$. Then, we have
$$f'(x)=\frac{1-\ln x}{x^2}\cdot x^{\frac 1x}.$$
So, $f(x)$ is increasing for $0\lt x\lt e$ and is decreasing for $x\gt e$.
With $\lim_{x\to 0^+}f(x)=0,\lim_{x\to\infty}f(x)=1$, we can say
$$x^{\frac 1x}\ge\frac 13\iff x\ge e^{-W(\ln 3)}\approx 0.548$$
where $W(x)$ is the Lambert W function with $f(e^{-W(\ln 3)})=1/3$.
By the way, note that $f(1/2)=1/4\lt 1/3$.
A: As long as $x^{1/x}>0$, your inequality is equivalent to
$$
\frac{\log x}{x} +\log 3 \geq 0. \tag{1}
$$
Take limits of the left-hand side of $(1)$ to prove that $(1)$ must be true for some $x$ sufficiently large. Then differentiate to prove that it must be true for all $x$ sufficiently large.
How do you prove that $b > 1/2$? Well, I guess you must do some numerical experiments... However, $(1/2)^2 = 1/4 < 1/3$.
A: Since of course $x$ must be positive, your inequality is equivalent to $$x\ge\left(\frac{1}{3}\right)^x,$$  whose LHS diverges while its RHS converges to $0$, so it holds for $x$ larger than some $b$. Since clearly $$\frac{1}{2}<\frac{1}{\sqrt{3}},$$ we conclude $$b>\frac{1}{2}.$$
