Prove: $1 + \frac12 x - \frac18 x^2 < \sqrt{1+x}, x > 0$

Question

To show $1 + \frac12 x - \frac18 x^2 < \sqrt{1+x}$ is it enough to tell that the taylor series expansion of $\sqrt{1+x}$ around $0$ has more positive terms?

Answer 1

If $x > 4$, then $LHS < 1 < \sqrt{x+1} = RHS$ and inequality holds. Assume $0 < x \le 4\implies LHS > 0$, square both sides and simplify:$$1+\dfrac{x^2}{4}+\dfrac{x^4}{64}+x- \dfrac{x^2}{4}- \dfrac{x^3}{8}< 1+x\iff \dfrac{x^4}{64}< \dfrac{x^3}{8}\iff\dfrac{x^3}{8}\left(\dfrac{x}{8}-1\right)< 0$$ which is true since $0 < x < 8$. The claim is verified...

Answer 2

Let $\varphi(x)=\sqrt{1+x}-1\dfrac{1}{2}x+\dfrac{1}{8}x^{2}$, $x\geq 0$, then for $x>0$, \begin{align*} \varphi(x)&=\varphi(x)-\varphi(0)\\ &=\varphi'(\xi)x\\ &=\left(\dfrac{1}{2}(1+\xi)^{-1/2}-\dfrac{1}{2}+\dfrac{1}{4}\xi\right)x. \end{align*} Now let $\eta(x)=\dfrac{1}{2}(1+x)^{-1/2}-\dfrac{1}{2}+\dfrac{1}{4}x$, $x\geq 0$, then for $x>0$, \begin{align*} \eta(x)&=\eta(x)-\eta(0)\\ &=\left(-\dfrac{1}{4}(1+\omega)^{-3/2}+\dfrac{1}{4}\right)x\\ &=\dfrac{1}{4}\left(1-\dfrac{1}{(1+\omega)^{3/2}}\right)x\\ &>0, \end{align*} so in particular, $\eta(\xi)>0$ because $0<\xi<x$, and hence $\varphi(x)>0$ for $x>0$.

Answer 3

From the integral form of the remainder for Taylor's theorem, $$\begin{split}(1+x)^{1/2}&=1+\tfrac{x}{2} -\tfrac{x^2}{8} +\tfrac{x^3}{16} \int_0^1 (1+tx)^{-5/2}3(1-t)^2\mathrm{d}t\\ &>1+\tfrac{x}{2} -\tfrac{x^2}{8} \end{split}$$ whenever $x>0$.

Answer 4

8 + 4x -x^2 < 8[1 + x]^.0.5

8 + x[ 4 - x] < 8[1 + x ]^0.5 , square both sides

64 + 16x[4-x] +x^2 [4 -x]^2 < 64[ 1 + x]

64 + 64 x - 16x^2 + x^2 [ x^2 -8x + 16 ] < 64 + 64x , cancel 64 + 64x

then - 16x^2 + x^4 - 8x^3 + 16x^2 < 0 , cancel - 16x^2 + 16x^2

x^4 - 8x^3 < 0

x^3[x - 8]< 0 is true when

x^3 < 0 AND [ x -8] > 0 that is when x < 0 and x > 8 , not feasable

OR

x ^3 > 0 AND [ x - 8] < 0 , that is when

0< x< 8 satisfies the inequality

Answer 5

The Maclaurin series of $\sqrt{1+x}$ is only convergent in a neighbourhood of the origin (due to the singularity at $x=-1$) hence it is not a good idea to use it for proving such inequality. Basic algebra performs much better: it is obvious that for $x>0$ we have $1+\frac{x}{2}>\sqrt{1+x}$ (it is enough to square both sides), and

$$0<1+\frac{x}{2}-\sqrt{1+x}=\frac{\left(1+\frac{x}{2}\right)^2-(1+x)}{1+\frac{x}{2}+\sqrt{1+x}}=\frac{x^2}{4\left(1+\frac{x}{2}+\sqrt{1+x}\right)} <\frac{x^2}{4(1+1)}=\frac{x^2}{8}.$$