Show that $\sqrt{1+x}0$ I am a little stuck on this question and would appreciate some help. The question asks me to prove that $\sqrt{1+x}&lt;1+\frac{x}{2}$ for all $x&gt;0$. 

I squared both sides of the question to get $1+x&lt;\frac{x^2}{4}+x+1$ for all $x&gt;0$. Then, I multiplied both sides by $4$ to get $4+4x&lt;x^2+4x+4$ for all $x&gt;0$. 

I am a little stuck and was wondering what to do after this step and how to actually provide sufficient proof to say that this statement is true.
You did well. Let's finish it:
we have $x^2&gt;0$ thus $\dfrac{x^2}{4}&gt;0$. Adding $x+1$ to both sides, we have
$$\dfrac{x^2}{4}+x+1&gt; x+1$$
or $$(\frac{x}{2}+1)^2&gt;x+1$$
or
$$\frac{x}{2}+1&gt;\sqrt{x+1}$$
The important principle here is that if $a,b$ are nonnegative real numbers, then $a&lt;b$ iff $a^2&lt;b^2.$ This property is used all the time. If we grant this, then all we need to show is that

$$(\sqrt{1+x})^2 &lt; (1+x/2)^2,$$

which is the same as saying $1+x &lt; 1+x + x^2/4.$ That is obviously true, and we're done.