<s>[INST] Prove $2^x≥x^2$ I could not understand this proof. How are we able to divide both sides of the inequality with different things and claim that the inequality stays the same? Plus, is this an induction proof?
If $x \in \mathbb{Z}$ and $x \ge 4$ then $2^x \ge x^2$ holds.
When $x = 4$ we have $2^4 = 16 \ge 16 = 4^2$.
We $x > 4$ we have $\frac{2^{x + 1}}{2^x} = 2$ and
$$
\frac{(x + 1)^2}{x^2}
= \left(1 + \frac{1}{x}\right)^2
\le \left(\frac{5x}{4}\right)^2
$$
since $2 \le 1.5625$
 [/INST] The inductive hypothesis is the line $2^x\ge x^2$ for $x\ge4$. Essentially what these steps suggest is the following:$$(x+1)^2\le(1.25x)^2\le2x^2\le2\cdot2^x=2^{x+1}$$which completes the induction proof.
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