Prove $2^x≥x^2$

Question

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 $2 \le 1.5625$

Answer 1

Yes , it is sort of an induction proof . We first show that the inequality holds for $x = 4$ . Now we find the ratios of the subsequent terms of $x^2$ and $2^x$ :

$$\dfrac{2^{x+1}}{2^x} = 2$$

And $$\dfrac{(x+1)^2}{x^2} = \left(1 + \dfrac{1}{x}\right)^2$$

Note that for $x\ge 4$ , the expression$\color{#d05}{\left(1 + \dfrac{1}{x}\right)^2 \le 2}$ . Since $2^x$ increases faster than $x^2$ , we can induct that $\color{#20f}{2^{x+1}\ge(x+1)^2}$for all $x\ge 4$ and the claim follows .

Answer 2

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.

Answer 3

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 $$\frac{x+1}x\le\sqrt2.$$