real analysis - Use the mean value to prove a certain result

I need to prove the following:

Use the Mean-Value Theorem to prove that:

$$\sqrt{1+h}<1+\frac{1}{2}h$$ for $h>0$

My attempt:

we first note that given that $h>0$ then

$$1+\frac{1}{2}h >1$$

and

$$1+h>1 \Rightarrow \sqrt{1+h}<\sqrt{1}=1$$

then we have that squaring both sides:

$$1+h<(1+\frac{1}{2}h)^{2}=1+h+\frac{1}{4}h^{2} \Rightarrow 0<\frac{1}{4}h^{2}$$

which is true, then we are done.

My question is How can I use the MVT to prove this? Thank you for your help.