Prove if $x > 0$, $1+\frac{1}{2}x-\frac{1}{8}x^2 \le \sqrt{1+x}$ Prove if $x > 0$,  $1+\frac{1}{2}x-\frac{1}{8}x^2 \le \sqrt{1+x}$.
I find online, one person suggested using Taylor Theorem to expand the right-hand side, and apply Bernoulli's inequality.
So, if $x_0 = 0$, $\sqrt{1+x} = 1+\frac{1}{2}x-\frac{1}{4(2!)}x^2+R$, where R is larger than $0$, this makes sense to me, but I'm trying to find another way to prove the equality. Like Mean Value Theorem for inequality $\sqrt{1+x} \le 1+\frac{1}{2}x$.
I see the part where $1+\frac{1}{2}x$ is the same, but get trouble to put $\frac{1}{8}x^2$ into the equation.

Well putting $x=\sinh^2(a)$
We have to show (using $\cosh^2(a)-\sinh^2(a)=1$):
$$\cosh(a)\geq 1+\frac{1}{2}\sinh^2(a)-\frac{1}{8}\sinh^4(a)$$
Or :
$$\sinh^6\Big(\frac{a}{2}\Big)(\cosh(a)+3)\geq 0$$
Wich is clearly obvious !
