Prove $2^x \ge\ x^2$ for all real $x \ge\ 4$. The above inequality can easily be proved by mathematical induction if the domain is $\mathbb{N}$ or $\mathbb{Q}$. How do we prove it if the domain becomes $\mathbb{R}$?
Although the result is obvious just by observing their graphs on a rectangular coordinate, I don't seem to be able to come up with a rigorous proof.
 A: Consider $f(x) = 2^x-x^2$, since $f(4)\ge0$ and $f$ is increasing there it follows.
To see that it's increasing you use the derivative $f'(x) = 2^x\ln 2 - 2x$. To see that this is positive you use that $f'(4)>0$ and $f'(x)$ is increasing, to see he later you just take the derivative again which is $f''(x) = 2^x(\ln2)^2-2$. Again we see that $f''(4)>0$ and increasing, again we see that its increasing because $f'''(x) = 2^x(\ln2)^3>0$.
Note about using induction which is definitely possible. If you've done this for $\mathbb Q$ you can use the fact that $2^x$ and $x^2$ is extended to $\mathbb R$ by requirin that $2^x$ and $x^2$ being continuous so then it follows that because $f(x)\ge 0$ for all rational $x\ge 4$ and that $f(x)$ is continuous it will follow that $f(x)\ge 0$ for all reals $x\ge 4$ too.
A: Hint : Logarithming we have $$x \ln(2)\ge 2 \ln(x),$$ so we have to show
       $$\frac{x}{\ln(x)} \ge \frac{2}{\ln(2)}$$ 
The inequalities are equivalent for $x\ge 4$ because of $\ln(x)>0$
Now, show that $f'(x)>0$ for $x\ge 4$, if we denote $f(x)=\frac{x}{\ln(x)}$.
Since $f(4)=\frac{2}{\ln(2)}$, the claim follows.
A: The inequality is the same as
$$
x\log 2\ge 2\log x
$$
(natural logarithm). Consider
$$
f(x)=x\log 2-2\log x
$$
(defined for $x>0$) so that
$$
f'(x)=\log 2-\frac{2}{x}
$$
that's positive for $x>2/\log2$. So the function is increasing in the interval $[2,\infty)$, because $2>2/\log 2$.
Since $f(2)=0$, …
A: $x^2$ is a parabolic curve which is horizontal while $2^x$ is a vertical ever increasing non linear curve expanding vertically. These graphs intersect at $(2)$ and the gradient for $2^x$ changes drastically. Going by the graphs is best way or we can do for induction.
