Jtdylktuy

$require{cancel}$

Show that the following inequality holds for $x>0$
$$1+frac{x}{2}-frac{x^2}{8}<sqrt{x+1}<1+frac{x}{2}.$$

I proceeded as follows

$$sqrt{x+1}=1+frac{x}{2}-frac{x^2}{8}+frac{x^3}{16 (xi+1)^{5/2}},quad xiin[0,x]$$
and substituing in the inequality yields
$$1+frac{x}{2}-frac{x^2}{8}<1+frac{x}{2}-frac{x^2}{8}+frac{x^3}{16 (xi+1)^{5/2}}<1+frac{x}{2}$$
which is equivalent to
$$begin{cases}
cancel{1+frac{x}{2}-frac{x^2}{8}}cancel{<1+frac{x}{2}-frac{x^2}{8}}+frac{x^3}{16 (xi+1)^{5/2}}quad(1)\
cancel{1+frac{x}{2}}-frac{x^2}{8}+frac{x^3}{16 (xi+1)^{5/2}}<cancel{1+frac{x}{2}}qquad,,,,,, (2)
end{cases}$$
Inequality $(1)$ holds for the values given by the problem statement. Instead for $(2)$
$$frac{x^3}{16 (xi+1)^{5/2}}<frac{x^2}{8}$$
But $$displaystyle{max_{0leqxileq x}Bigg{frac{1}{16 (xi+1)^{5/2}}Bigg}}=displaystyle{min_{0leqxileq x}{Big{16 (xi+1)^{5/2}}Big}}$$
which occurs at $xi=0$, and thus, for $(2)$, it is left to prove that
$$frac{x^3}{16}<frac{x^2}{8}$$
but this happens for $x<0,vee,0<x<2$.

Instead, if I use $xi=x$ then inequality $(2)$ becomes
$$frac{x^3}{16 (x+1)^{5/2}}<frac{x^2}{8}$$
which holds for $-1<x<0,vee,x>0$, and thus coincides with the restriction given by the problem.

Would it be correct to take $xi=x$ rather than $0$? Is this approach correct at all?

The right inequality:

We need to prove that:
$$1+x<left(1+frac{x}{2}right)^2$$ or
$$1+x<1+x+frac{x^2}{4}.$$
The left inequality.

Let $x+1=t^2,$ where $t>1$.

Thus, we need to prove that:
$$1+frac{t^2-1}{2}-frac{(t^2-1)^2}{8}<t$$ or
$$(t-1)^3(t+3)>0.$$

At $x=0$, all three members equal $1$. Then we can take the derivative and

$$frac12-frac x4<frac1{2sqrt{x+1}}<frac12.$$

Again, we have equality at $x=0$ and

$$-frac14<-frac1{4(x+1)^{3/2}}<0.$$

This final bracketing is obvious.

For the left inequality:
$$
1 + frac{x}{2} - sqrt{1 + x} = frac{left(1 + frac{x}{2}right)^2 - (1 + x)}{1 + frac{x}{2} + sqrt{1 + x}} < frac{frac{x^2}{4}}{2} = frac{x^2}{8}.
$$

Well the inequality can be proved using algebra by squaring it. A proof via Taylor's theorem is as follows. We have a number $cin(0,x)$ such that $$sqrt {1+x}=1+frac{x}{2}-frac{x^2}{8}(1+c)^{-3/2}$$ The desired inequality follows because $0<(1+c)^{-3/2}<1$.

Your approach is similar but unnecessarily uses third derivatives and complicates the situation.