## May 2000#2

Problem Statement: Use the Mean Value Theorem to show that $\sqrt{x+1}<1+\dfrac{x}{2}$ for all $x>0$.

Proof: Let $f(x)=1+\dfrac{x}{2}-\sqrt{x+1}$. We wish to show that $f(x)>0$ for every $x>0$. First note that $f'(x)=\dfrac{1}{2}-\dfrac{1}{2\sqrt{x+1}}>0$ for all $x>0$. Let $x>0$ and consider the interval $[o,x]$. Since $f(x)$ is continuous and differentiable on $[0,x]$ we may apply the Mean Value Theorem. According to the MVT there exists a point $c\in(0,x)$ such that $f'(c)=\dfrac{f(x)-f(0)}{x-0}$. We know that $f(0)=0$ so $f'(c)=\dfrac{f(x)}{x}$. Since we know $f'(x)>0$ the previous statement implies that $0. Since $x$ was an arbitrary number greater than $0$ we may conclude that $f(x)>0$ for every $x>0$.

$\Box$

Reflection: This one actually took me a while. My first thought was to use the function $f(x)$ as defined above, but instead of looking at the interval $[0,x]$ I looked at an arbitrary interval $[a,b]$. This allowed me to conclude that $f(a)-f(b)>0$ but that didn’t give me what I wanted. It was after realizing that $f(0)=0$ and that I could use $0$ as an endpoint of the interval the proof kind of fell out.