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.
 A: 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.
A: Using CS-inequality: let $x = a+b+c \implies \left(x+\frac{19}{2}\right)^2 = \left(1\sqrt{a+1}+2\sqrt{b+2}+3\sqrt{c+3}\right)^2\leq 14(x+6)\implies \left(x+\frac{5}{2}\right)^2 \leq 0\implies x = -\dfrac{5}{2}$
A: You can also compute this on a complete direct approach without cool formulas.
For this, you first need to solve it for one variable, for example $c$. Then you may insert the solution for $c$ and solve the next, $b$. Finally you can solve $a$ and receive an independent value for it.
Next you put in the solution for $a$ in the dependent solution for $b$ and get an independent value for $b$, put this in $c$ and viola, you are finished.
The hard part is to solve this for $c$, you will need to take everything to the power of $2$ for getting rid of the squares.
Well, have fun with $($loooong thing$)^2$ :)
Wolfram Alpha might be handy for such stuff: Your query at Wolfram Alpha
