Prove a generalization of Bernoulli's inequality

For real numbers $a_1, \ldots, a_n$ with $a_i > -1$ for all $i = 1, \ldots, n$ and all of the $a_i$ having the same sign, prove

$(1+a_1)(1+a_2) \cdots (1+a_n) \geq 1 + a_1 + a_2 + \cdots + a_n.$

As a special case let $a_1 = \cdots = a_n = x$ and prove Bernoulli’s inequality,

$(1+x)^n \geq 1 + nx.$

Finally, show that if $n > 1$ then equality holds only when $x = 0$ .

Proof. The proof is by induction. For the case $n =1$ , we have,

$(1+a_1) = 1 + a_1 \qquad \implies \qquad 1+a_1 \geq 1+a_1,$

so the inequality holds for $n=1$ .
Assume then that the inequality holds for some $n = k \in \mathbb{Z}_{>0}$ . Then,

$\begin{align*} &&(1+a_1) \cdots (1+a_k) &\geq 1 + a_1 + \cdots + a_k \\ \implies&& (1+a_1) \cdots (1+a_k)(1+a_{k+1}) &\geq (1+a_1 + \cdots + a_k)(1+a_{k+1})\\ &&&\geq (1+a_1+ \cdots + a_{k+1})+a_{k+1}(a_1 + \cdots + a_k). \end{align*}$

But, $a_{k+1}(a_1 + \cdots + a_k) \geq 0$ since every $a_i$ must have the same sign (thus, $a_{k+1}$ and $(a_1 + \cdots + a_k)$ must have the same sign, so the product is positive). Thus,

$(1+a_1) \cdots (1+a_{k+1}) \geq 1+a_1 + \cdots + a_{k+1}.$

Hence, the inequality holds for the case $k+1$ ; and therefore, for all $n \in \mathbb{Z}_{>0}. \qquad \blacksquare$

Now, if $a_1 = \cdots = a_n = x$ where $x > -1$ and $x \neq 0$ we apply the theorem above to obtain Bernoulli’s inequality,

$(1+a_1) \cdots (1+a_n) = (1+x)^n \geq 1+a_1 + \cdots + a_n = 1+n \cdot x.$

Claim: Equality holds in Bernoulli’s inequality if and only if $x = 0$ .
Proof.
If $x = 0$ then $(1+x)^n = 1 = 1 + n \cdot x$ , so indeed equality holds for $x = 0$ . Next, we use induction to show that if $x \neq 0$ , then the inequality must be strict. (Hence, equality holds if and only if $x = 0$ .)
For the case $n=2$ , on the left we have,

$(1+x)^2 = 1+2x + x^2 > 1+2x$

since $x^2 > 0$ for $x \neq 0$ . So, the inequality is strict for the case $n =2$ . Assume then that the inequality is strict for some $n = k \in \mathbb{Z}_{>1}$ . Then,

$\begin{align*} (1+x)^k > 1+kx\ &\implies & (1+x)^k (1+x) &> (1+kx)(1+x) &(x>-1 \implies 1+x>0)\\ &\implies & (1+x)^{k+1} &> 1 + (k+1)x + kx^2 \\ &\implies & (1+x)^{k+1} &> 1+(k+1)x. \end{align*}$

Where the final line follows since $k > 0$ and $x > 0$ implies $kx^2 > 0$ . Therefore, the inequality is strict for all $n > 1$ if $x \neq 0$ .

Hence, the equality holds if and only if $x = 0. \qquad \blacksquare$

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Prove a generalization of Bernoulli’s inequality

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