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$$ \sin (\sqrt{1+x}) < \dfrac{1}{2}x + \sin 1 $$ 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 $\dfrac{\cos (\sqrt{1+x})}{2\sqrt{x+1}}$.

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  • $\begingroup$ I can't help but suspect that the binomial series approximation $\sqrt{1+x}\simeq1+\dfrac x2$ is somehow relevant to this question. $\endgroup$ – Lucian Oct 12 '15 at 4:48
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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 $f(x)\gt 0$.

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And $\cos\sqrt{1+x}\le1$ and $\sqrt{1+x}>1$ for $x>0$. Which combined results in the claim.

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