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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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