calculus - How can I prove $log_2{(1+x)} geq {x}$ for $0

How can I prove the following inequality for $0 $$\log_2{(1+x)} \geq {x}$$

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.

Answer

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 $[0,1]$.