<s>[INST] 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.
 [/INST] We’ll let $a = x + 1$
Transform the inequality by substitute $x$ with $a - 1$.
After some calculations we the get
$$ -a^2 + 6a + 3 \leq 8\sqrt{a} $$
Square both sides and we get
$$ a^4 - 12a^3 + 30a^2 - 28a + 9 \leq 0$$
Which is just $(a-1)^3(a - 9) \leq 0$.
Since $a > 1$ , this statement is true for all $a \in [1,9]$.
Thus, we are left to prove this statement for all $a > 9$.
Since $a = x + 1$ , $x > 8$.
Consider $(x - 2)^2 + 12 > 0$
After some calculations we then get $\frac{1}{2}x < 3  - 1 + \frac{1}{8}x^2$.
But since $x > 8$ we then get $1 + \frac{1}{2}x - \frac{1}{8}x^2 < \sqrt{x + 1}$.
Hence, this inequality is true for all $x > 0$.
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