Show using Bernoulli inequality approach (using induction, & transitivity) $2^n \ge n^2, \ \ \forall n\ge 4, n\in \mathbb{N}$.

Question

It is given in the book by John B. Reade, titled: Introduction to Mathematical Analysis; on pg. 7; to prove the inequality

$$2^n \ge n^2, \ \ \forall n\ge 4, n\in \mathbb{N}.$$

It states that has distilled the proof using induction, & transitivity to the proof for the inequality: $2n^2\ge (n+1)^2, \forall n \ge 4$.

Am confused & feel that the proof before it for showing (Bernoulli inequality) $(1+x)^n\ge 1+nx$, by using induction, & transitivity must be applicable. Although, not clear how. Am presenting my attempt below based on that proof, but am unable to get the desired form of the inequality using that.

The inequality is true for the base case: $n=(n_0=)4$, i.e. $2^4 \ge 4^2$ is true.

Assume it is true for a particular $n\gt 4$ (inductive step), need show that it implies it is also true for $n+1$.

$2^{n +1} = 2^n.2\ge (n+1)^2\implies 2^n.2\ge (n+1)^2 = 2^{(n+1)}\ge (n+1)^2$. This can't be pursued further.

2. Also, need show that $2^n.2\ge 2n^2$; but that leads to another proof using induction, & transitivity, as shown below. (I assume that the author takes this inequality to be known, hence not elaborates it).

For $n \ge 4, n\in \mathbb{N}$, need show $2^n.2\ge 2n^2$.
Base case $n=4: 2^4.2 \ge 2.16$.
Inductive case assumed for a value $n$.
Need show that $2^{(n+1)}.2\ge 2(n+1)^2\implies 2^{(n+1)}\ge (n+1)^2$

So, going nowhere.

Answer 1

Clearly, we don't really need $(1 + x)^n \ge 1+nx$ to prove that $2n^2 \ge (n+1)^2, \forall n \ge 4$.

Since it is just saying that $$n^2 \ge 2n+1, \forall n \ge 4$$

which is equivalent to $$(n-1)^2 \ge 2, \forall n \ge 4$$

which we know it is true since if $n \ge 4$, the $(n-1)^2 \ge 9 \ge 2$.

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If we insist to use $(1+x)^n \ge 1+nx$, we have

$$\left( 1-\frac1n\right)^2 \ge 1-\frac2n$$

Note that for $n \ge 5$, we have $1-\frac2n \ge 1-\frac25=\frac35 \ge \frac12 $.

Hence for $n \ge 5$, we have $\left(1-\frac1n\right)^2 \ge \frac12.$

which is equivalent to $$ 2(n-1)^2 \ge n^2, \forall n \ge 5$$

which is equivalent to

$$2n^2 \ge \left( 1+n\right)^2, \forall n \ge 4.$$