Find $a + b + c$, given that $(a+1)^{1/2} - a + (b+2)^{1/2} \cdot 2 - b + (c+3)^{1/2} \cdot 3 - c = \frac{19}{2}.$ 
Let $a,b,c$ be real number such that $$(a+1)^{1/2} - a + (b+2)^{1/2} \cdot 2 - b + (c+3)^{1/2} \cdot 3 - c = \frac{19}{2}.$$
  Find $a + b + c$.

The answer is: $-\frac{5}{2}$. Please give me some clues or solution. 
Thanks.

The function $(a+1)^{1/2}-a$ has derivative $\frac{1}{2(a+1)^{1/2}}-1$, so reaches a max of $5/4$ at $a=-3/4$.
A similar calculation shows that $2(b+2)^{1/2}-b$ reaches a max of $3$ at $b=-1$, and the function $3(c+3)^{1/2}$ reaches a max of $21/4$ at $c=-3/4$. 
Add up. The sum is $19/2$. What a coincidence! Thus $a+b+c=(-3/4)+(-1)+(-3/4)$.
Another way: In hindsight, it could have been done less messily. For example, let $b+2=y^2$. Then we are looking at $2y-(y^2-2)$, which is maximized at $y=1$. Similarly with the others.
