interpreting $(1+x)^n \ge 1 + nx (1 + \tfrac{x}{2})^{n-1} \quad (x \ge 0, \; n \in \mathbb{N})$

Question

I'd like to understand the simple relation $$ (1+x)^n \ge 1 + nx (1 + \tfrac{x}{2})^{n-1} \qquad (x \ge 0, \; n \in \mathbb{N}) $$ pictorially. To this end I rewrote it as $$ \frac{(1+x)^n -1}{x} \ge n(1+\tfrac{x}{2})^{n-1} = \frac{d}{dt} (1+t)^n \bigg|_{t=\frac{x}{2}} $$ and examined slopes pertaining to the graph of $(1+t)^n$. The left member is the slope of the chord joining the points at $t=0$ and $t=x$; by the mean value theorem for derivatives, this must equal the slope of the tangent line at some point $\xi$ between $0$ and $x$. The right member, on the other hand, is the slope of the tangent line at the midpoint $t=\frac{x}{2}$. So the inequality seems to assert that $\xi$ must actually fall to the right of $\tfrac{x}{2}$. This seems intuitively reasonable (since the local slope is increasing with $t$) but does it follow from the mere convexity and increasing nature of $(1+t)^n$?

Answer 1

No, it doesn't follow, because $(1+t)^n$ is convex and increasing for $n=3/2$, too, though the inequality is not true for $n=3/2$ and $x=1$. It seems to be crucial that $(1+t)^{n-1}$ is still convex. At least, that's sufficient:
Claim If $f'(x)$ is convex on $[a,b]$, then $$\frac{f(b)-f(a)}{b-a}\ge f'\left(\frac{a+b}2\right)\tag1.$$
Proof: $$\frac{f(b)-f(a)}{b-a}=\int^1_0f'(a+(b-a)t)\,dt=\int^1_0f'(b-(b-a)t)\,dt,$$ and the arithmetic mean of both integrands is $$\frac{f'(a+(b-a)t)+f'(b-(b-a)t)}2\ge f'\left(\frac{a+b}2\right)$$ by convexity of $f'$. qed
Of course, $x^{n-1}$ is convex for every integer $n\ge1$, so (1) is valid for $f(x)=x^n$, letting $a=1$ and $b=1+x$ answers the question.