# Unicity of solution of $\left(1+x\right)^\alpha = 1+\alpha x$

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:

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).

• It's Bernoulli's inequality ! 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 $$\alpha\neq 0,1$$

• Your answer is useful, but I look explicitly to a solution without differentiation Jan 26 at 13:42