How prove that:$\varphi(2)+\varphi(3)+\varphi(4)+\cdots+\varphi(n)\ge\frac{n(n-1)}{4}+1$

In the following picture the black dots represent the grid points (below the line $x=y$) that are visible from the origin, ie those for which there is no other grid point in the segment joining the point and the origin.

Within the triangle there are exactly $\varphi(k)$ black dots in the $k$th column, say those with coordinates $(a,k)$ with $\gcd(a,k)=1$ and $1 \le a \le k$, as a consequence the number of black dots in a triangle $T(n)$ with corners $(2,1),(n,1)$ and $(n,n-1)$ (including the borders) is exactly $$ B(n) = \varphi(2) + \varphi(3) + \dots + \varphi(n)$$ Call $W(n)$ the number of white dots in the triangle then $$ W(n) = 1-\varphi(2) + 2-\varphi(3) + \dots + n-1-\varphi(n) = \frac{n(n-1)}2 - B(n)$$

Given $n>3$ Consider the two triangles $T(n)$ and $T(\lfloor n/2 \rfloor)$, the second is represented with the green area, and the first with yellow and green areas.

We are going to show that the number of white dots in the big triangle $T(n)$ is at most $$ 2 B(\lfloor n/2 \rfloor) + 2 W(\lfloor n/2 \rfloor) = \lfloor n/2 \rfloor(\lfloor n/2 \rfloor-1) $$ we consider the lines going throug the origin and a white dot we have the following facts:

1) every such line cointains exacly a black dot, and it is in the green area.

2) the number of grid points in the line inside the yellow area is at most one more than the number of grid points in the line inside the green area.

As a consequence the number of white dots in the yellow area is at most equal to $B(\lfloor n/2 \rfloor)+W((\lfloor n/2 \rfloor)$ (the total number of points in the green triangle) plus $B(\lfloor n/2 \rfloor)$ the total number of lines going through white dots). Adding the number of white dots in the green triangle we obtain that the totla number of white dots in the big triangle is bounded by

$$2 (B(\lfloor n/2 \rfloor) + W(\lfloor n/2 \rfloor) = \lfloor n/2 \rfloor((\lfloor n/2 \rfloor-1) $$

So the number of black dots in the big triangle verifies:

$$B(n) \ge \frac{n(n-1)}2 - \lfloor n/2 \rfloor((\lfloor n/2 \rfloor-1) $$

if $n\ge 4$ then the right hand side is at least $$ \frac{n(n-1)}2 - \frac n2 \left(\frac n2-1\right) = \frac{n^2}4 \ge \frac{n(n-1)}4 + \frac{n}{4} \ge \frac{n(n-1)}4 + 1 $$ as was to be proved.

EDIT: I have rewritten the proof as the original was very confusing and contained some incorrect statements.