Maximum value for $M^4$ 
$$M=|\sqrt{x^2+4x+5}-\sqrt{x^2+2x+4}|$$
  Find the maximum value of $M^4$.

I think it could have a geometric solution. Because it looks like the difference between two points formula. Please help.

Assuming $x \in \mathbb{R}$, 
$$M(x) = \left \lvert \sqrt{x^2 + 4 x + 5} - \sqrt{x^2 + 2 x + 4} \right \rvert$$
and
$$M(x)^4 = (M(x)^2)^2 = g(x)^2$$
where
$$g(x) = \left ( \sqrt{x^2 + 4 x + 5} - \sqrt{x^2 + 2 x + 4} \right )^2$$
As Jaideep Khare and Abishanka Saha already succintly answered, we can write
$$M(x) = \left \lvert \sqrt{(x - 2)^2 + (0 \pm 1)^2} - \sqrt{(x - 1)^2 + (0 \pm \sqrt{3})^2} \right \rvert$$ 
where we can choose each $\pm$ between $+$ or $-$ without affecting the result. If we choose arbitrarily
$$M(x) = \left \lvert \sqrt{(x - 2)^2 + (0 - 1)^2} - \sqrt{(x - 1)^2 + (0 - \sqrt{3})^2} \right \rvert$$ 
then we can immediately see, as aforementioned members already explained, that $M(x)$ is the difference in distances from $(x,0)$ to $(2,1)$ and $(x,0)$ and $(1,\sqrt{3})$, which allows us to describe a result based on geometric reasoning.
Personally, as a non-mathematician, I felt unsure which triangle inequality would apply here, and was too lazy to find out. We have a triangle, with $(x, 0)$ at one vertex, and $(2,\pm1)$ and $(1,\pm\sqrt{3})$ as the other two, arbitrarily choosing either $+$ or $-$ for each $\pm$ to get the maximum difference in the length of edges meeting at $(x,0)$. 
I get unreasonably suspicious — paranoid — whenever I need to make arbitrary decisions without a clear rule as to exactly why an arbitrary decision is necessary here, as opposed to some other rule stating which choice should be taken. I'm a physicist, and the first thing I always examine is whether the model I applied makes sense; and here, I just could not tell offhand why it should.
So, I decided to tackle it as a non-geometric math problem, and find out.
We know that a real-valued function $g(x)$ with $x \in \mathbb{R}$, reaches its (local) extrema whenever its derivative is zero. The global maximum is one of those. For the derivative in this case, we need to apply the chain rule (fixed!),
$$\frac{d b(a(x))}{d x} = \left [ \frac{d b(y)}{d y}\right ]_{y = a(x)} \frac{d a(x)}{d x}$$
twice. The subscript of the bracket means that after derivation with respect to $y$, we substitute $y$ with $b(x)$. In this case, the outermost function is square, the inner function is a square root, and the innermost are the functions inside the square roots in $g(x)$. We get
$$\frac{d g(x)}{d x} = 2 \left ( \sqrt{x^2 + 4 x + 5} - \sqrt{x^2 + 2 x + 4} \right ) \left ( \frac{2 x + 4}{2 \sqrt{x^2 + 4 x + 5}} - \frac{2 x + 2}{2 \sqrt{x^2 + 2 x + 4}} \right )$$
$$= 2 \left ( \sqrt{x^2 + 4 x + 5} - \sqrt{x^2 + 2 x + 4} \right) \left (\frac{ (x + 2)\sqrt{x^2 + 2 x + 4} - (x + 1)\sqrt{x^2 + 4 x + 5} }{\sqrt{(x^2 + 4 x + 5)(x^2 + 2 x + 4)}}\right)$$
This reaches zero when either part in parentheses is zero; the latter part in parentheses is only zero if and only if the numerator is zero:
$$\frac{d g(x)}{d x} = 0 \; \iff \; \begin{cases}
\sqrt{x^2 + 4 x + 5} - \sqrt{x^2 + 2 x + 4} = 0 & \text{or} \\
(x + 2)\sqrt{x^2 + 2 x + 4} - (x + 1)\sqrt{x^2 + 4 x + 5} = 0 \end{cases}$$
The upper one is zero when $x = -1/2$, and the lower one when $x = -5/2 - \sqrt{3}/2$. Since $g(-1/2) = 0$ (and $g(x) \ge 0$ for all $x \in \mathbb{R}$, since $(y)^2 \ge 0$ for all $y \in \mathbb{R}$), it is the (or a) minimum; and since we only have two extrema (and $g(x) \ne 0$ for at least some $x \in \mathbb{R}$), $g(-5/2-\sqrt{3}/2)$ must be the global maximum.
Therefore, the maximum value of $M(x)^4 = g(x)^2$ when $x = -5/2 - \sqrt{3}/2$.
It was interesting (and reassuring!) to note that $g(-5/2-\sqrt{3}/2)$ corresponds to the solution obtained via geometric interpretation, $\left(\sqrt{(2-1)^2 + (1 - \sqrt{3})^2}\right)^2$.  If we were to examine why this is so, and explore other similar problems, we'd onclude that the reason is we're governed by the basic triangle inequality here — which also explains the hint in Jaideep Kahre's answer: it is the special equality case, maximum, in the triangle inequality rule.
(I like that: it ties the loose ends I was unsure about in this problem very nicely. I often check my math this way, too — although I usually use e.g. Maple to do the hard work for me. I just pick the models that I think apply to this case, and see if they agree. If they do not, I definitely erred in choosing the model, and must examine where my understanding failed. If they agree, I've examined the problem from at least two different viewpoints, and therefore have at least some confidence in my solution. This is also the reason I posted this answer, even though there are already two concise answers posted: I wanted to show how I would approach the stated problem, as a weak-math-fu non-mathematician.)
The actual maximum value is obtained with straightforward substitution:
$$M \left(\frac{-5}{2} - \frac{\sqrt{3}}{2}\right)^4 = g \left(\frac{-5}{2} - \frac{\sqrt{3}}{2}\right)^4 = \left[\sqrt{\left(-\frac{5}{2} - \frac{\sqrt{3}}{2}\right)^2 - 5 - 2 \sqrt{3}} - \sqrt{\left(-\frac{5}{2} - \frac{\sqrt{3}}{2}\right)^2 - 1 - \sqrt{3} }\right]^8$$
$$g \left(\frac{-5}{2} - \frac{\sqrt{3}}{2}\right) = \left( 5 - 2 \sqrt{3} \right)^4 = 2569 - 1480 \sqrt{3} \approx 5.564804$$
