<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] If $$2^x\ge x^2$$ is known to hold, then
$$2^{x+1}\ge(x+1)^2$$ because
$$2^{x+1}=2\cdot2^x\ge 2\cdot x^2\ge\left(\dfrac{x+1}{x}\right)^2x^2=(x+1)^2$$ provided $$\frac{x+1}x\le\sqrt2.$$
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