# Closest-point heuristic approximation problem

I have an Approximation Algorithm problem and its answer but I am having a hard time to clearly understand.
The problem is as follows:

Consider the following closest-point heuristic for building an approximate traveling-salesman tour whose cost function satisfies the triangle inequality. Begin with a trivial cycle consisting of a single arbitrarily chosen vertex. At each step, identify the vertex $u$ that is not on the cycle but whose distance to any vertex on the cycle is minimum. Suppose that the vertex on the cycle that is nearest $u$ is vertex $v$. Extend the cycle to include $u$ by inserting $u$ just after $v$. Repeat until all vertices are on the cycle. Prove that this heuristic returns a tour whose total cost is not more than twice the cost of an optimal tour.

The answer is (I apoligize for including an image but I couldn´t get the some special characters displayed correctly) What I do not understand is the part surrounded by the red rectangle. I will very much appreciate if you help me to get the idea behind it. I am studying for an exam and I am stuck with this problem.

Respectfully,
$$c(H_i)\leq c(H_{i-1})+2c(u_i, v_i)$$
Let's say initially it was $$u\to x$$. Then you add $$v$$ directly after $$u$$, forming $$u\to v\to x$$. This increases the total distance by $$c(u,v)+c(v,x)-c(u,x)$$. From the triangle inequality, we have $$c(u,v) + c(u,x) \geq c(v,x)$$ and thus $$2c(u,v)\geq c(u,v)+c(v,x)-c(u,x)$$, hence $$c(H_i)\leq c(H_{i-1})+2c(u_i, v_i)$$.