Alternative proof of $x^x \geq \sin x$ if $x>0$

Question

My lecturer gave me this exercise: "Show that for $x>0$ it is $x^x \geq \sin x$."

This is my approach: since $x \geq \sin x$ for all $x \geq 0$, it is sufficient to show that $x^x \geq x$; since $x>0$ we can divide by $x$ and find out that $x^x \geq x \iff x^{x-1} \geq 1 \iff e^{(x-1) \log x} \geq 1 \iff (x-1) \log x \geq 0$ and this is true for all $x>0$.

However I was curious if another approach of the following kind is possible: define $D(x):=x^x-\sin x$, so the problem is equivalent to show that $D(x)>0$ for all $x>0$; since $D(x)\to \infty$ for $x \to \infty$ and $D(x)\to1$ for $x\to0^+$ we know that $$\forall M>0 \ \exists K_M \ \text{s.t.} \ x> K_M \implies D(x)>M$$ $$\forall\varepsilon>0 \ \exists \delta_{\varepsilon}>0 \ \text{s.t.} \ 0<x<\delta_{\varepsilon} \implies1-\varepsilon<D(x)<1+\varepsilon$$ So for the arbitrarity of $\varepsilon>0$ and $M>0$, we can deduce that in the interval $(0,\delta_\varepsilon)\cup(K_M,\infty)$ it is $D(x)>0$.

So, if I'm not wrong, proving the inequality is equivalent to prove that $K_M<\delta_{\varepsilon}$; but I don't know if this is possible, since we know that those numbers exist but we don't know them and they depend on $M,\varepsilon$.

Thank you.

Answer 1

The inequality is obvious for $x> 1$, so we restrict to $[0,1]$. It suffices to show $D(x)$ is convex, as convex functions have a unique global minimum, i.e. it suffices to show $D''(x)>0$. Calculating we have $$ D''(x) = x^{x-1}+x^x(1+\log(x))^2+\sin(x)>0 $$Using Newton's Method or the bisection method or some such, the unique real root of $D'$ is $x_0\approx 0.820453$, with $D(x_0)\approx 0.118675$.