$2^x\leq x+1$ for $x\in [0,1]$ I tried using mean value theorem but couldn't show $2^{2^c} < e$ for $c ∈ (0,x)$. Writing taylor expansion of $2^x$ also don't work because we need to show $2^x$ is smaller than something, not greater
 A: It's equivalent to asking $2^x-x-1\leq 0$ for $x\in{0;1}$. Define $f(x) = 2^x - x -1$ and it's derivative is $f'(x) = \ln(2)2^x-1$. You can see the derivative is negative for $x\leq x_{min}\approx 0.53$ and postive for $x\geq x_{\min}$.
Also, $f(0) = f(1) = 0$.
A monotonic function on a closed interval has both maximum and minimum in the borders of the intevals. So, for $x\in[0;x_{\min}]$, $f$ is decreasing, $f(0) = 0$ and $f(x_{\min})\leq 0$ therefore, it's maximum is $0$. Then, it's negative on that inteval. Same argument for $x\in[x_{\min},1]$.
A: Here are a few approaches for proving this inequality and a graphical demonstration.

Convexity
Since the second derivative is everywhere positive:
$$
\begin{align}
\frac{\mathrm{d}^2}{\mathrm{d}x^2}2^x
&=\log(2)^22^x\\
&\gt0\tag1
\end{align}
$$
the function $2^x$ is convex.
If $f$ is convex on $[a,b]$, then for $x\in[a,b]$,
$$
f(x)\le f(a)\frac{b-x}{b-a}+f(b)\frac{x-a}{b-a}\tag2
$$
Setting $f(x)=2^x$, $a=0$, and $b=1$, $(2)$ says that for $x\in[0,1]$,
$$
\begin{align}
2^x
&\le2^0\frac{1-x}{1-0}+2^1\frac{x-0}{1-0}\\
&=1+x\tag3
\end{align}
$$

Graph


Mean Value Theorem
Suppose that $f$ is convex on $[a,b]$. The Mean Value Theorem then says
$$
\frac{f(x)-f(a)}{x-a}=f(\xi_a)\quad\text{for some $\xi_a\in(a,x)$}\tag4
$$
and
$$
\frac{f(b)-f(x)}{b-x}=f(\xi_b)\quad\text{for some $\xi_a\in(x,b)$}\tag5
$$
If $f''(x)\ge0$, then $f'(x)$ is increasing on $[a,b]$. Therefore, since $\xi_a\lt x\lt\xi_b$, we have $f'(\xi_a)\le f'(\xi_b)$, and thus $(4)$ and $(5)$ say
$$
\frac{f(x)-f(a)}{x-a}\le\frac{f(b)-f(x)}{b-x}\tag6
$$
which becomes
$$
f(x)\le f(a)\frac{b-x}{b-a}+f(b)\frac{x-a}{b-a}\tag7
$$
Thus, the Mean Value Theorem shows that $f''(x)\ge0$ implies $(2)$.

Bernoulli's Inequality
For $x\in[0,1]$, Bernoulli's Inequality is reversed:
$$
(1+a)^x\le1+ax\tag8
$$
Plugging in $a=1$ yields
$$
2^x\le1+x\tag9
$$
A: Start with $2^x=1+x$ has only two roots in the interval namely $0$ and $1$.  Therefore $2^x \gt 1+x$ for the entire open interval or $2^x\lt 1+x$ for the entire open interval, since both sides are continuous.  Look at the point $x=.5$.  There $2^x=\sqrt{2}=1.414...$, while $1+x=1.5$, so $2^x\lt 1+x$ at this point - thus for the open interval $(0,1)$
A: Suppose $f(x)=1+x-2^x$, $x \in [0,1]$.
From $f^{\prime}(x)=1-2^x \cdot \log 2$ we have $f$ $\uparrow$ on $[0,c]$ and $\downarrow$ on $[c,1]$, where $2^c=1/\log 2$. Note that $0<c<1$.
Since $f(0)=0=f(1)$, we have $f \ge 0$ on $[0,1]$. $\blacksquare$
A: $$2^x-x-1 = 0$$ only when $x=0$ or $x=1$ on the interval. So if there exists a point in between such that the function is greater than $0$, then the whole interval must be greater than $0$ by the intermediate value theorem.
A: According to the final bullet point under properties of functions of one variable at the Wikipedia entry on convex functions, a continuous function is convex if it is "midpoint convex," where midpoint convexity means
$$f\left(x+y\over2\right)\le{f(x)+f(y)\over2}$$
for all $x$ and $y$.
For $f(x)=2^x$, midpoint convexity amounts to the familiar (and easily proved) arithmetic-geometric mean inequality
$$\sqrt{uv}\le\left(u+v\over2\right)$$
with $u=2^x$ and $v=2^y$. Thus $2^x$ is convex, hence lies on or below the line connecting $(0,1)$ and $(1,2)$ on the interval $[0,1]$, i.e., $2^x\le x+1$ for $0\le x\le1$.
Remark: The theorem that midpoint convexity implies convexity (for continuous functions) is intuitively obvious if you draw a picture showing repeated bisections of an interval $[a,b]$ converging to a point $x$ in its interior.
A: Since other people have explained the typical way to do this, I'll mention that there's a very tricky way to show $2^x \leq 1 + x$ for $0 \leq x \leq 1$ with an infinite series expansion. But not a Taylor series - an infinite series of binomial coefficients.
If we define the binomial coefficients $\binom{x}{k}$ for non-integer arguments $x$ as $$\binom{x}{0} := 1 \text{ and } \binom{x}{k} := \frac{x(x-1)...(x-k+1)}{k!} \text{ for } k \geq 1,$$
then $$2^x = \sum_{k = 0}^\infty \binom{x}{k}$$ is an absolutely convergent series for $x > 0$, and a conditionally convergent series for $0 > x > -1$†, as we have the asymptotic formula $$\binom{x}{k} \approx \frac{(-1)^k}{\Gamma(-x)k^{x+1}}$$ as $k \rightarrow \infty$. Writing out the first few terms explicitly:
$$2^x = \sum_{k = 0}^\infty \binom{x}{k} = 1 + x + \binom{x}{2} + \binom{x}{3} + ...$$
We are now ready to get our inequality $2^x \leq 1 + x$ for $0 \leq x \leq 1$. In fact, we can prove something stronger: $2^x < 1 + x$ for $0 < x < 1$ (i.e. the inequality is strict everywhere except the endpoints).
For $0 < x < 1$ and $k \geq 1$, the terms $\binom{x}{k}$ are nonzero, alternating in sign, and their absolute values are monotone decreasing towards zero, since $$\binom{x}{k+1} = \frac{x-k}{k+1} \binom{x}{k} = (-1)\frac{k-x}{k+1} \binom{x}{k},$$
and $0 < \frac{k-x}{k+1} < 1$ when $k \geq 1$ and $0 < x < 1$.
Therefore, in the range $0 < x < 1$, $$1 + x + \binom{x}{2} + \binom{x}{3} + ...$$ satisfies the conditions of the alternating series test from the second term onwards. Because in an alternating series, the odd partial sums are upper bounds and the even partial sums are lower bounds when the first term (in this case, $1+x$) is positive, we get a whole family of inequalities for $2^x$, among them the one in the original post:
\begin{align*}
2^x & \leq 1 + x \\
2^x & \geq 1 + x + \binom{x}{2} \\
2^x & \leq 1 + x + \binom{x}{2}+\binom{x}{3} \\
2^x & \geq 1 + x + \binom{x}{2}+\binom{x}{3}+ \binom{x}{4} \\
 & ... \\
2^x & \leq 1 + x + \binom{x}{2}+...+\binom{x}{2n-1} \\
2^x & \geq 1 + x + \binom{x}{2}+...+\binom{x}{2n} \\
 &... \
\end{align*}
for all $0 \leq x \leq 1$ and all $n \geq 1$, with equality only if $x = 0$ or $x = 1$.

† When $x$ equals $0$, it is a finite series: $\binom{0}{0} = 1$ and $\binom{0}{k} = 0$ for all other $k$, so $2^0 = \sum_{k = 0}^\infty \binom{0}{k} = 1 + 0 + 0 + ... = 1.$ This is true more generally: for $x$ any nonnegative integer, the series will be finite, with all but its first $x+1$ terms equal to $0$.
