How to use other method to prove $2^x>2x-1$.

let $x\in R$,show $$2^x>2x-1$$

My try: take a function $f(x)=2^x-2x+1$, then we have $$f'(x)=2^x\ln{2}-2,~~~~f''(x)=2^x\ln^2{2}>0$$ if let $x_{0}$ such $f'(x_{0})=0\Longrightarrow 2^{x_{0}}=\dfrac{2}{\ln{2}}$ is minimum the point (since $f''(x)>0$).

so we have

$$f(x)\ge f(x_{0})= \dfrac{2}{\ln{2}}-1+\dfrac{\ln{\ln{2}}}{\ln{2}}+1=\dfrac{2+\ln{\ln{2}}}{\ln{2}}>0$$ because $$2+\ln{\ln{2}}>0\Longleftrightarrow 2^{e^2}>e$$ have other methods?Thanks

Solution 1:

Let's see if we can prove the result with resorting to logs or calculus.

We'll begin with a rewrite of $2^x\gt2x-1$, letting $x=2u+{1\over2}$, as

$$4^u\gt2\sqrt2u$$

Since the inequality is obvious for $u\le0$ (since $u^u$ is always positive), it suffices to prove the stronger inequality

$$u\le{4^u\over3}$$

for $u\ge0$. Toward this end, it will help to know that

$${3\over2}\lt4^{1/3},\quad2=4^{1/2},\quad{5\over2}\lt4^{2/3},\quad3\lt4^{5/6},\quad4=4^1,\quad\text{and}\quad6\lt4^{4/3}$$

(The first three inequalities amount to ${27\over8}\lt4$, ${125\over8}\lt16$, and $27\lt32$; the last inequality essentially repeats the first.) We now see that

$$\begin{align} 0\le u\le{1\over3}&\implies u\le{4^0\over3}\le{4^u\over3}\\ {1\over3}\le u\le{1\over2}&\implies u\le{3/2\over3}\lt{4^{1/3}\over3}\le{4^u\over3}\\ {1\over2}\le u\le{2\over3}&\implies u\le{4^{1/2}\over3}\le{4^u\over3}\\ {2\over3}\le u\le{5\over6}&\implies u\le{5/2\over3}\lt{4^{2/3}\over3}\le{4^u\over3}\\ {5\over6}\le u\le1&\implies u\le{3\over3}\lt{4^{5/6}\over3}\le{4^u\over3}\\ 1\le u\le{4\over3}&\implies u\le{4^1\over3}\le{4^u\over3}\\ {4\over3}\le u\le2&\implies u\le{6\over3}\lt{4^{4/3}\over3}\le{4^u\over3} \end{align}$$

This establishes the inequality for all $u\in[0,2]$. To show it for $u\gt2$, it's enough to prove that $n+1\le4^n/3$ for all $2\le n\in\mathbb{N}$, since that inequality tells us that for $u\ge2$,

$$u\le\lfloor u\rfloor+1\le{4^{\lfloor u\rfloor}\over3}\le{4^u\over3}$$

The proof of $n+1\le 4^n/3$ is by induction: the inequality is easily checked for the base case $n=2$ after which we see that

$$n+1\le{4^n\over3}\implies(n+1)+1\le{4^n\over3}+1\le{4^n\over3}+{3\cdot4^n\over3}={4^{n+1}\over3}$$

And this completes the proof of $2^x\gt2x-1$.

Remark: I did not expect the proof to require as many incremental steps (breaking $[0,2]$ into seven pieces) as it turned out to use. I'd be keen to see an alternative that uses fewer.

Solution 2:

For $x<\frac{1}{2}, LHS>0>RHS.$

For $\frac{1}{2}<x<1, LHS>\sqrt{2}>1>RHS$.

For $1<x<2$, $$2^x = e^{x\ln 2}$$ $$=\sum_{i=0}^\infty\frac{(x\ln2)^i}{i!}$$ $$>1+x\sum_{i=1}^\infty\frac{(\ln2)^i}{i!}$$ $$=1+x(e^{\ln2}-1)$$ $$=1+x$$ $$>2x-1$$ For $x\geq 2$, $$2^x\geq2x>2x-1$$ Equality holds at $x=2$, and further rate of growth (i.e. derivative) of LHS is more than that of RHS. So, we are done.

Solution 3:

The following 'argument' is guided by following a geometric/intuitive path. Every claim can be justified using calculus, but it would be of interest to see how much of that firepower is really required.

We have two functions,

$\tag 1 f(x) = 2^x$ $\tag 2 g(x) = 2x - 1$

To say that the OP's inequality is false is equivalent to saying the graphs of these two functions have a nonempty intersection. This is equivalent to finding a $k \le -1$ and a linear function,

$\tag 3 h(x) = 2x + k$

such that the graph of $h(x)$ intersects the graph of $f(x)$ at exactly one point.

Let $(x_0,y_0)$ be this point of intersection.

Using calculus/algebra it can be shown that

$\tag 4 x_0 = \displaystyle{\frac{\ln(2) - \ln(\ln(2))}{\ln(2)}}$

The value of $k$ is the $y\text{-intercept}$, and so, using algebra,

$\tag 5 k = 2^{x_0} - 2 x_0 \approx -0.172 $

But this contradicts the premise that $k \le -1$. In conclusion, the graph of a function of the form $\text{(3)}$ is disjoint from the graphs of $f(x)$ if and only if:

$\tag 6 k \lt 2^{x_0} - 2 x_0$