How to prove $x^{x/(1+x)}/(1+x)\geq1/2$

Question

I need to show that $\frac{x^{\frac{x}{1+x}}}{1+x} \geq \frac{1}{2}$ for all $x\geq0$.

I am fairly certain this statement is true, but have no idea how to go about proving it.

I know that it attains a local minimum of 1/2 at $x=1$. It also seems to hold when I plot it out, decreasing from $[0,1]$ and increasing after, it seems to approaches 1 as $x\rightarrow\infty$.

Answer 1

By Bernoulli $$\frac{x^{\frac{x}{x+1}}}{1+x}=\frac{1}{(1+x)\left(1+\frac{1}{x}-1\right)^{\frac{x}{x+1}}}\geq\frac{1}{(1+x)\left(1+\left(\frac{1}{x}-1\right)\cdot\frac{x}{x+1}\right)}=\frac{1}{2}.$$

Answer 2

If we call this function $f(x)$ then since $f(1)=\frac12$ you must have intended to say that $f(x)\geq\frac12$ in your statement.

Additionally $f(x)$ does have a local minimum (not local limit, whatever that means) of $\frac12$ at $x=1$ and is increasing for $x>1$. This follows from finding $f'(x)$ and finding that $f'(1)=0$ even though it is a bit messy.

That is all you need.

Graphing the function shows it approaches $1$ not $2$ as $x$ goes to $\infty$. To prove this, L'Hopital's rule did not seem to work so perhaps someone else can fill this in by other means but it's not needed to prove your claim.

Added: as often happens, the above answer appeared just as I posted this!

Answer 3

$$f(x)=\frac{x^{\frac{x}{x+1}}}{x+1}\quad \implies \quad f'(x)=\frac{x^{\frac{x}{x+1}}}{(x+1)^3} \log (x)$$

The first derivative cancels only once at $x=1$.

$$f(1)=\frac 12 \qquad f'(1)=0\qquad f''(1)=\frac 18 $$

So, this is a minimum.

When $x$ is large $$f(x) \sim 1-\frac{\log (x)+1}{x}$$ which is the curved asymptote.

Edit

For building the asymptotics, write $$\log(f(x))=\frac x{x+1}\log(x)-\log(x+1)$$ Using Taylor expansion $$\log(f(x))=-\frac{\log (x)+1}{x}+\frac{2 \log (x)+1}{2 x^2}+O\left(\frac{1}{x^3}\right)$$

Taylor again $$f(x)=e^{\log(f(x)) }=1-\frac{\log (x)+1}{x}+\frac{\log ^2(x)+4 \log (x)+2}{2 x^2}+O\left(\frac{1}{x^3}\right)$$