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.

The inequality can be proved using elementary algebra. If $\dfrac{1}{8}x^2 < 1+\dfrac{1}{2}x\implies 2-2\sqrt{3} <x < 2+2\sqrt{3} < 8$, then both sides are positive, and square both of them: $\left(1+\dfrac{1}{2}x-\dfrac{1}{8}x^2\right)^2 < 1+x\iff \left(1+\dfrac{1}{2}x\right)^2-2\left(1+\dfrac{1}{2}x\right)\left(\dfrac{1}{8}x^2\right)+\left(\dfrac{1}{8}x^2\right)^2< 1+x\iff -x+\dfrac{1}{8}x^2 < 0\iff \dfrac{1}{8}x^2 < x\iff x^2 - 8x < 0\iff x(x-8) < 0\iff 0 < x < 8$, and this is true since we've shown that $8 > x > 0$, and if $\dfrac{1}{8}x^2 \ge 1+\dfrac{1}{2}x$, then it is obviously true since the left side is non-positive and the right side is non-negative. You're done !
