Elementary proof of Bernoulli's inequality for rational exponents

Question

I was wondering if anyone has or can come up with a proof of $(1+\alpha)^q \geq 1+q\alpha$ with $q>1$ a rational number and $\alpha >0 $ a real number. Bernoulli's inequality is easily proven for natural number exponents using induction and general real exponents using derivatives. I'd like to see if there is an elementary proof that just uses the field axioms and immediate results from them, simple arithmetic, etc. In other words, I just don't want to use differentiation and beyond. Arguments with sequences and convergence would be fine, but I'd still prefer just straight up algebra/arithmetic. Also, we can let $\alpha \geq -1$, but I'm just concerned with positive $\alpha$. Presumably once I have the argument for $\alpha>0$, then it will be easy to extend to allowable negative values. Similarly I'll worry about $0<q<1$ later assuming it is a simple extension of the result for $q>1$.

Let $\alpha>0$ and $n,k\in\mathbb N$ with $q=\frac{n+k}{n}>1$ be the rational exponent we are considering. It's easy to show that $(1+\alpha)^q=1+\delta$ for some $\delta\in\mathbb R$ such that $\delta>\alpha$ and $(1+\alpha)^{n+k}=(1+\delta)^n$ (with rational exponentiation defined precisely in that way).

I have come up with the equation: $$(1+\alpha)^{\frac{n+k}{n}}=1+\frac{n+k}{n}\alpha \cdot \left[\frac{\frac{1}{n+k}\frac1\alpha\left((1+\alpha)^{n+k}-1\right)}{\frac{1}{n}\frac1\delta\left((1+\delta)^{n}-1\right)}\right].$$ So my thought is to show that the bracketed expression is greater than one. Note that since $(1+\alpha)^{n+k}=(1+\delta)^n$, the expression on the right just simplifies to $1+\delta$ as desired. So it's not some kind of strange result.

I spent some time looking at the $n=2,k=1$ case $(1+\alpha)^\frac32$, but am just stumped and think maybe I'm just venturing down a rabbit hole. Or maybe need to seek more expert input before I waste any more time with this.

Answer 1

You don't need any high-powered machinery to show this. If $q = m/n > 1$, then after raising both sides of our desired inequality to the $n$th power, we get $$(1 + \alpha)^m \geq (1 + q\alpha)^n.$$ Expanding both sides out binomially, and comparing coefficients of $\alpha$, it becomes clear that to demonstrate the inequality for $\alpha > 0$, we just need to check that for any pair of integers $n < m$ and any integer $0 \leq k \leq n$, $$\binom{m}{k} \geq \frac{m^k}{n^k}\binom{n}{k}.$$ This is clearly true when $k = 0$. Otherwise, $$\frac{\binom{m}{k}}{\binom{n}{k}} = \frac{m(m-1)...(m-k+1)}{n(n-1)...(n-k+1)} \geq \frac{m^k}{n^k},$$ since $$\frac{m-i}{n-i} \geq \frac{m}{n}$$ for $0 \leq i < n$.

[The last inequality $\frac{m-i}{n-i} \geq \frac{m}{n}$ is proven by cross-multiplying: $$(m-i)n = mn - ni \geq mn - mi = (n-i)m.]$$

Answer 2

Select $p$ to be the Hoelder conjugate, i.e. since $q\gt 1$ we select positive rational $p$ such that $\frac{1}{q} +\frac{1}{p} = 1$.

take $q$th roots of each side and prove the equivalent
$\big(1+q\alpha\big)^\frac{1}{q}$
$=\big(1+q\alpha\big)^\frac{1}{q} \cdot 1^\frac{1}{p}$
$\leq \frac{1}{q}\cdot \big(1+q\alpha\big) +\frac{1}{p}\cdot 1$
$= 1+\alpha$
by $\text{GM}\leq \text{AM}$ which is easily proven over the rationals by e.g. Cauchy's forward-backward induction.

Answer 3

Let $q = \frac{m}{n}$ with $m > n$. By AM-GM, we have $$(1 + q\alpha)^{1/q} = \sqrt[m]{(1 + q\alpha)^n \cdot 1^{m-n}} \le \frac{(1+q\alpha)n + 1\cdot (m-n)}{m} = 1 + q\alpha \frac{n}{m} = 1 + \alpha.$$ We are done.

Answer 4

Let me show a quick proof for any real $r\geq1$ and $x\geq-1$.

Consider a function $f(x)=(1+x)^r$. It's easy to check that it's a convex function when $r\geq1$ and $x\geq-1$. We will use the fact that convex function lies above any of its tangent lines. That is $f(x)\geq f(x_0)+f'(x_0)(x-x_0)$ for any $x$ and $x_0$. In particular $f(x)\geq f(0)+f'(0)x$.

In our case this would lead to Bernoulli inequality.

Let's prove the above statement. From the definition of convextiy $$\alpha f(x)+(1-\alpha)f(x_0)\geq f(\alpha x +(1-\alpha)x_0) \qquad \forall x,x_0 \text{ and } \alpha\in(0,1)$$ By rearranging we get $$\frac{f(x)-f(x_0)}{x-x_0}\geq \frac{f(x_0 +\alpha(x-x_0))}{\alpha(x-x_0)}$$ This is true for any $x$,$x_0$ and $\alpha\in(0,1)$. In particular if $\alpha\to 0^+$. Note that if we denote by $h = \alpha(x-x_0)$ this will also tend to 0. Thus $\frac{f(x_0 +h)}{h}\to f'(x_0)$ which proves our inequality.