How prove that:$\varphi(2)+\varphi(3)+\varphi(4)+\cdots+\varphi(n)\ge\frac{n(n-1)}{4}+1$ Prove that for $n\ge 3$,
$$\varphi(2)+\varphi(3)+\varphi(4)+\cdots+\varphi(n)\ge\dfrac{n(n-1)}{4}+1$$
where $\varphi$ is the Euler's totient function
I think we must use this
$$\sum_{k=1}^{n}\varphi(k)=\dfrac{1}{2}\left(1+\sum_{k=1}^{n}\mu(k)\left[\dfrac{n}{k}\right]^2\right)$$

This is only a partial answer, just a possibly profitable approach. 
On the rhs we have an expression which contains just the half of the sum of consecutive numbers (with a small deviation), so I would try the problem comparing the double of the sum of totients with that of the natural numbers.
The conjecture is
$$ \varphi(2) + \varphi(3) + ... + \varphi(n) \ge {n(n-1)\over 4} + 1 = {n(n+1)\over 2} \frac12 - \frac n2+1 $$
The $\varphi(\cdot)$ and the expression for the sum-of-consecutive-numbers expanded gives
$$
2\cdot(1-\frac12) + 3\cdot(1-\frac13)+4\cdot(1-\frac12)+...+ \varphi(n) \ge (1+2+3+4+...+n)(1-\frac12) - \frac n2 +1 $$
and we see, that most of the parentheses on the lhs evaluate to more than $(1-\frac12)=\frac12$
To simplify multiply both sides by 2 to get
$$
(2 + 3 \cdot\frac43+4 + 5 \cdot\frac85 + 6 \cdot\frac 23  + 7 \cdot\frac {12}7 +...+2 \varphi(n)) \ge (1+2+3+4+...+n) -  n +2 $$ 
Then we get first
$$(2 + (3 +1)+4 + (5 +3)+ (6 - 6/3)  + (7 + 5)+...+2 \varphi(n)) \ge (1+2+3+4+...+n) -  n +2 $$ and separating the lhs in two parentheses
$$ \begin{eqnarray}  && (2 + 3 +4 + 5 + 6  + 7+...+ n) \\
 &+ & (0+1+0+3-2+5+0+3... + 2 \varphi(n)-n ) \\ &\ge & (1+2+3+4+...+n) & -  n +2 \end{eqnarray} $$
and finally 
$$  0+1+0+3-2+5+0+3 + \ldots + 2\varphi(n)  \ge 3  $$
The problem reduces then to show, that the partial sums of the lhs-expression are always positive and $\ge 3$
