Prove: $x^2<2^x$, $\forall x\ge x_0$, $x\in \Bbb R$

Question

Prove: $\exists x_0\in \Bbb R$ such $\forall x\in \Bbb R$, $x\ge x_0$, it is true that $x^2<2^x$.

My attempt: if $n\in \Bbb N$ the problem would be trivial, as it is not difficult to prove that $n^2<2^n$, for $n\ge 5$. How can I find an appropriate argument here with $x\in \Bbb R$ ? Hints and answers are welcomed.

Answer 1

Let $[x]=n$.

Thus, there is $n\geq3$, for which $$2^x\geq2^n=(1+1)^n\geq1+n+\frac{n(n-1)}{2}+\frac{n(n-1)(n-2)}{6}>(n+1)^2>x^2.$$ My previous solution was tautological and I deleted it.

Answer 2

The easiest way is probably to differentiate, to show that the difference $2^x - x^2$ is increasing for $x > x_0$; then you just need to show that $2^x - x^2$ is greater than $0$ at $x_0$.

Answer 3

Abusing $(n+a)^2<2(n^2+a^2)$ we have that if $x=n+a$ for $0\leq a<1$ and $n\in \Bbb{N}$ then it's sufficient to prove $$2^{n+a}=2^n\cdot 2^a>2(n^2+a^2)\\2^{n-1}\cdot 2^a>n^2+a^2$$ You can even take a stronger inequality to eliminate the $a$'s $$2^{n-1}\cdot 2^a\geq2^{n-1}>n^2+1>n^2+a^2$$ So you're left to prove $2^{n-1}>n^2+1$ for all integer $n$ bigger or equal then some $n_0$ (you can easily find a suitable $n_0$)

Answer 4

Personally I would turn to the definition of $2^x$ where $x \in \mathbb{R}$, which is defined as $2^x := \exp(x \ln(2))$.

Then by definition we have:

$$2^x = \sum_{n=0}^\infty \frac{x^n \ln(2)^n}{n!}$$

Then whenever $x \neq 0$, we have: $$\frac{2^x}{x^2} = \sum_{n=0}^\infty \frac{x^{n-2} \ln(2)^n}{n!}$$ $$\frac{2^x}{x^2} = \frac{1}{x^2} + \frac{\ln(2)}{x} + \sum_{n=0}^\infty \frac{x^n \ln(2)^{n+2}}{(n+2)!}$$

The first two terms on the right-hand side tend to $0$ as $x$ tends to $+\infty$, whilst the third term tends to $+\infty$.

Then by the arithmetic of limits, we have $\lim_{x \to +\infty} \frac{2^x}{x^2} = +\infty$.

Therefore there exists $x_0 \in \mathbb{R}$, such that for all $x \geq x_0$ we have $\frac{2^x}{x^2} > 1$. In other words $2^x > x^2$ for all $x \geq x_0$.