How to prove $\ln(1+x)\ge x/(1+x/2)$ when $x > 0$? I think the equation establishes only when $x = 0$, so the condition should be x>=0. The first derivative of $\ln(1+x)-x/(1+x/2)$ is $0$ at $x=0$. So the inequality looks good to establish. Is there a better way to prove it?

To show
$\ln(1+x)
\ge x/(1+x/2)
$.
Since
$\ln(1+x)
=\int_0^x \dfrac{dt}{1+t}
$,
this is equivalent to
$\dfrac1{x}\int_0^x \dfrac{dt}{1+t}
\ge \dfrac1{1+x/2}
$.
Since
$\dfrac1{1+x/2}$
is the value of
$\dfrac1{1+t}
$
at the midpoint,
I'll combine the two halves.
$\begin{array}\\
\int_0^x \frac{dt}{1+t}
&=\int_0^{x/2} \frac{dt}{1+t}+\int_{x/2}^x \frac{dt}{1+t}\\
&=\int_0^{x/2} \frac{dt}{1+t}+\int_0^{x/2} \frac{dt}{1+x-t}\\
&=\int_0^{x/2} dt\left(\frac{1}{1+t}+\frac{1}{1+x-t}\right)\\
&=\int_0^{x/2} dt\left(\frac{1+x-t+1+t}{(1+t)(1+x-t)}\right)\\
&=\int_0^{x/2} dt\left(\frac{2+x}{1+t+x-t+t(x-t)}\right)\\
&=(2+x)\int_0^{x/2} dt\left(\frac{1}{1+x+t(x-t)}\right)\\
&\ge(2+x)\int_0^{x/2} dt\left(\frac{1}{1+x+x^2/4}\right)
\qquad\text{since }t(x-t) \le x^2/4\\
&=\dfrac{2+x}{(1+x/2)^2}\int_0^{x/2} dt\\
&=\dfrac{x}{2}\dfrac{4(2+x)}{(2+x)^2}\\
&=\dfrac{2x}{x+2}\\
&=\dfrac{x}{1+x/2}\\
\end{array}
$
so that
$\dfrac1{x}\int_0^x \frac{dt}{1+t}
\ge \dfrac{1}{1+x/2}
$.
Note that
the only inequality used
is that
$t(x-t)
\le x^2/4$
for $0 \le t \le x/2$.
If we use
$t(x-t) \ge 0$,
we can get an inequality
going the other way:
$\begin{array}\\
\int_0^x \frac{dt}{1+t}
&=(2+x)\int_0^{x/2} dt\left(\frac{1}{1+x+t(x-t)}\right)\\
&\le(2+x)\int_0^{x/2} dt\left(\frac{1}{1+x}\right)\\
&=\dfrac{2+x}{1+x}\int_0^{x/2} dt\\
&=\dfrac{x(2+x)}{2(1+x)}\\
\end{array}
$
so that
$\ln(1+x)
\le \dfrac{x(2+x)}{2(1+x)}
$.
The difference of the two bounds is
$\begin{array}\\
\dfrac{x(2+x)}{2(1+x)}-\dfrac{2x}{x+2}
&=\dfrac{x(2+x)(x+2)-2x(2(1+x))}{2(1+x)(x+2)}\\
&=\dfrac{x(x^2+4x+4-(4+4x))}{2(1+x)(x+2)}\\
&=\dfrac{x^3}{2(1+x)(x+2)}\\
&\le \dfrac{x^3}{4}\\
\end{array}
$
