Mean value theorem to show that $\root\of{1+4x} < \frac{5+2x}{3}$ for all $x\gt3$. Recalling the theorem; for some $f:[a,b]\rightarrow\mathbb R$ be a continuous function that is differentiable on $(a,b)$ then there is some $c\in(a,b)$ such that 
$$f(b)-f(a) = f'(c)(b-a)$$
I assume the approach to this proof assumes that these two function are differentiable on $[3,\infty)$.
I'm wondering if someone could give out a hint as to how to approach this problem so I can learn from it. Thankyou.
 A: The trick here is to show the inequality for $x>2$ instead of $x>3$. So, apply the MVT to the interval $[a,b] = [2,x]$. Set


*

*$f(x) = \sqrt{1+4x}$

*$\frac{f(x) - f(2)}{x-2} = f'(\xi)$ for a $\xi \in (2,x)$
$$\Rightarrow \frac{\sqrt{1+4x} - 3}{x-2} = \frac{2}{\sqrt{1+4\xi}}\stackrel{\xi>2}{<}\frac{2}{3}\Rightarrow \color{blue}{\sqrt{1+4x}\stackrel{x>2}{<}} \frac{2}{3}(x-2) + 3  = \color{blue}{\frac{5+2x}{3}}$$
A: So I followed a great resource here; (https://www.youtube.com/watch?v=dO4zbUTcf_M)
Anyhow, we're interested in the interval $(3,x)$ and we know that the function $\root\of{1+4x}$ (call if $f$) is indeed differentiable on this interval (I hope this assumption is fine, I can't see why not). We can apply MVT in the natural way: 
$$\frac{f(x) - f(3)}{x-3}=\frac{\root\of{1+4x} - \root\of{1+(4*3)}}{x-3} = f'(c)=\frac{2}{\root\of{1+4c}} \lt \frac{2}{\root\of{1+(4*3)}}$$
By simple manipulation, we find this equation is leads to:
$$\root\of{1+4x} \lt \frac{2x+7}{\root\of{13}}$$
But it's not exactly a hit just yet.
A: Hint: Show that $\frac{5}{3}+\frac{2x}{3}$ is the tangent line of 
$\sqrt{1+4x}$ at $x= \underline{\text{fill in the blank}}$. Next note that $\sqrt{1+4x}$ is strictly concave down for $x>0$. Then make your conclusion.
