calculus - Inequality using mean value theorem

$$\sin (\sqrt{1+x}) < \dfrac{1}{2}x + \sin 1$$ for $x > 0$.

Judging from the chapter this exercise is given in, I'm guessing you can do this using the mean value theorem. I don't get how though. I understand the theorem, but how can I apply it to this inequality?

I calculated that the derivative of $\sin (\sqrt{1+x})$ is $\dfrac{\cos (\sqrt{1+x})}{2\sqrt{x+1}}$.

Answer

Let $f(x)=\frac{x}{2}+\sin 1-\sin(\sqrt{1+x})$. Then $f(0)=0$, and
$$f'(x)=\frac{1}{2}+\frac{1}{2\sqrt{1+x}}\cos(\sqrt{1+x}).$$
Note that $f'(x)\gt 0$ for all $x\gt 0$. So $f$ is increasing on $(0,\infty)$.

Remark: If we want to use the MVT directly, suppose $x\gt 0$. Note that $f(x)=f(x)-f(0)$, and

$$\frac{f(x)-f(0)}{x-0}=f'(\xi)$$ for some $\xi$ between $0$ and $x$. But, as noted above, $f'(\xi)\gt 0$. So $f(x)-f(0)\gt 0$, and therefore $f(x)\gt 0$.