Let \(h(x)=\sqrt{1+x}-(1+\frac{1}{2}x)\text{.}\) The desired inequality is equivalent to the inequality \(h(x)< 0\) for all \(x> 0\text{,}\) which we now endeavor to demonstrate. First observe that \(h'(x)< 0\) for all \(x> 0\text{.}\) Indeed, we have
\begin{align*}
h'(x) \amp = \frac{1}{2\sqrt{1+x}}-\frac{1}{2}\\
\amp < \frac{1}{2}-\frac{1}{2} \amp (x> 0\implies \sqrt{1+x}> 1)\\
\amp = 0\text{.}
\end{align*}
Next, since \(h\) is continuous on and differentiable on \([0,\infty)\text{,}\) given any \(x> 0\text{,}\) the mean value theorem implies that there exists a \(c\in (0,x)\) satisfying
\begin{equation*}
h'(c)=\frac{h(x)-h(0)}{x-0}=\frac{h(x)-0}{x}=\frac{h(x)}{x}\text{.}
\end{equation*}
This implies
\begin{equation*}
h(x)=xh'(c)< 0\text{,}
\end{equation*}
since \(x> 0\) and \(h'(c)< 0\text{.}\) We have shown that \(h(x)< 0\) for all \(x>0 \text{,}\) and hence that \(\sqrt{1+x}< 1+\frac{1}{2}x\) for all \(x> 0\text{.}\)