# Use Lagrange Remainder Theorem to Prove Inequality

I'm supposed to use Lagrange Remainder Theorem to prove that

$$1 + \frac{x}{2} - \frac{x^2}{8} < \sqrt{1+x} < 1 + \frac{x}{2} \text{ } \text{ if } x>0$$

Obviously, the left and right hand sides look like Taylor series, but how would you do this using Lagrange, it isn't obviously intuitive. Thanks!

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I've added links to mathworld and wikipedia, I hope you don't mind. (This might be useful, since not everybody must know the theorem or someone might know it, but not under this name.) – Martin Sleziak Dec 2 '11 at 8:25

For the RHS inequality you can even use Mean Value Theorem (which, of course, is a special case of Taylor theorem - where the reminder term contains first derivative).

Let $f(x) = \sqrt{1+x}$. Then the theorem gives that there exists some $\theta \in (0,x)$ such that $$\frac{f(x)-f(0)}{x-0} = \frac{\sqrt{1+x} - 1}{x} = f'(\theta) = \frac{1}{2 \sqrt{1+\theta}}.$$ Hence $\sqrt{1+x} = 1 + \frac{x}{2 \sqrt{1+\theta}} < 1 + \frac{x}{2}$ since $x > 0$. Observe that, of course, $\frac{x}{2 \sqrt{1+\theta}}$ is the reminder term in the Taylor theorem.

For the LHS by Taylor theorem you get that there exists $\theta \in (0,x)$ such that $$\sqrt{1+x} = 1 + \frac{1}{2} x - \frac{1}{8} x^2 + \frac{1}{16} (1 + \theta)^{-5/2} x^3.$$ The reminder term is positive for $x > 0$, which proves the inequality.

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Hint Let $f(x)= \sqrt{1+x}$ and let $T_n$ be the nth degree Taylor Polynomial.

Then $R_n(x) = f(x)-T_n(x)$.

Now, what happens if you can prove that some $R_n(x)$ is positive or negative?

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