<s>[INST] Prove that $\sqrt{(1+x)}<1+\frac{1}{2}x$ $\forall x>0$ I was just practicing some real analysis for an upcoming quiz and I just wanted some feedback on if my proof was written sufficiently. The question's above, my answer's below. Thanks!
Pf: Let $x>0,$ $ x \in (0,x)$, and let $f(x) = \sqrt{(1+x)}-(1+\frac{1}{2}x)$. Observe that $f(0) = \sqrt{1+0} - (1 + \frac{1}{2}0) = 1 - 1 = 0.$ Also observe that $f'(x) = \frac{1}{2\sqrt{1+x}} - \frac{1}{2}.$ We know that $\forall x>0$, $\sqrt{1+x}>1$ implies that $\frac{1}{2\sqrt{1+x}} - \frac{1}{2} < 0.$ Thus $f(x)$ is decreasing by the Decreasing Criterion and $f(x) < 0 \Rightarrow$ $\sqrt{(1+x)}-(1+\frac{1}{2}x) < 0 \Rightarrow$  $\sqrt{(1+x)}<1+\frac{1}{2}x$ $\forall x>0.$
 [/INST] Let $x>0$, then $\frac{1}{4}x^2>0$
By adding $1+x$ to both sides, we obtain:
$$1+x+\frac{1}{4}x^2>1+x$$It follows that:$$(1+\dfrac{1}{2}x)^2>1+x$$
Since both sides are positive for all $x>0$, then $$\sqrt{(1+x)}<1+\dfrac{1}{2}x$$
</s>