How can I prove the following inequality for $0<x<1$ $$\log_2{(1+x)} \geq {x}$$
I tried to use $\ln(1+x) \geq x-\frac{x^2}{2}$ for $x\geq 0$ and convert the $\ln$ to $\log_2$ to prove that, but it does not work. Any idea? Is there a generalized form?
I need an analytical proof, not sketch , or looking at graph. Thanks.
On
Hint: Exponentiate the inequality to the equivalent form $1+x\ge2^x$, and sketch the two graphs $y=1+x$ and $y=2^x$. Note that $(x,y)=(0,1)$ and $(1,2)$ are common to both graphs.
On
Consider the first derivate :$(\frac{\ln(1+x)}{\ln2})' = \frac{1}{\ln2(1+x)} > 0$ of $0<x<1$, so our function is increase. $x$ also incease, so. We should just check what will be on one point. For example $x = 3$ second function is bigger, so on $0 < x < 1$ the first one will be bigger than second (because of they intersects in $x = 1$).
On
The key is, the derivative of the function $f(x)=\log_{2}(1+x)-x$ is $g(x)=\log_2 e\cdot\frac{1}{1+x} - 1$. The function $g(x)$ is decreasing for $x\in [0, 1]$, and the value is from positive to negative. So the minimum of $f(x)$ is at $0$ and $1$.
On
As Barry said, convert to the equivalent form $1+x \geq 2^x$. Then, consider the function $f(x)=1+x-2^x$. Taking two derivatives you get that $f''(x)=-2^x (\ln(2))^2$ which is always negative. Hence $f$ is concave. Since $f(0)=0=f(1)$, it follows that $f$ must be positive on $(0,1)$, which is what you wanted.
The logarithm function is concave.
Thus, it lies above any secant line.
In particular, on $[0,1]$, the function $x$ is a secant line for $\log_2(1+x)$.
Hence, $\log_2(1+x)\ge x$ on $[0,1]$.