Consider the two Bernoulli (thank you @ErikSatie) inequalities for $x>-1$: $$ 0<\alpha<1 \quad \to\quad \left(1+x\right)^\alpha\le 1+\alpha x,\tag{1} $$ $$ \alpha<0\lor 1<\alpha \quad \to\quad \left(1+x\right)^\alpha\ge 1+\alpha x.\tag{2} $$

It remains me to prove that the inequality is strict for every $x\ne 0$ (keeping $x>-1$)

It seams to me that this point is more difficult than the inequalities themselves. I can't use derivation.

In the following picture:

enter image description here

I plotted different cases of $\alpha\in\{-1,-0.1,0.1,0.5,0.9,1.1,2\}$, and with a different color the asymptotic cases $\alpha=0$ (yellow) and $\alpha=1$ (green).

What I have to do is to prove that in $(-1,\infty)$ there is only a single solution for $$ \left(1+x\right)^\alpha = 1+\alpha x,\quad \alpha\in (-\infty,0)\cup(0,1)\cup (1,\infty).\tag{3} $$

EDIT 1 The case (1) derives directly for the corresponding aspect for the arithmetic-geometric inequality (which is used to prove it).

  • $\begingroup$ It's Bernoulli's inequality ! $\endgroup$
    – Erik Satie
    Jan 3 at 13:33

Consider the function $f(x) = (1+x)^{\alpha}-1-\alpha x$. We know that $f(0)=0$, and we essentially wish to show that for $\alpha\neq 0,1$, $x=0$ is the only zero of $f(x)$.

Firstly, we can easily see that $f(x)$ is twice differentiable on $\mathbb{R}/\{0\}$ with first derivative $f'(x) = \alpha(1+x)^{\alpha-1}-\alpha$ and second derivative $f''(x)=\alpha(\alpha-1)(1+x)^{\alpha-2}$.

From this, we can see that $f'(0)=0$, and $f''(x)\neq0$ except at $\alpha=0,1$ where it is zero throughout.

Now, if there exists another $x_0\neq 0$ at which $f(x_0)=0$, then by Rolle's theorem, we would have a $c$ between $x_0$ and $0$ such that $f'(c)=0$.

But, then as $f'(0)=0$, we can again apply Rolle's theorem to get a $d$ between $c$ and $0$ at which $f''(d)=0$. This is impossible, unless $\alpha=0,1$.

Hence, the Bernoulli inequality is strict for every $x\neq 0$ for $\alpha\neq 0,1$

  • $\begingroup$ Your answer is useful, but I look explicitly to a solution without differentiation $\endgroup$
    – Wyatt
    Jan 26 at 13:42

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