<s>[INST] 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.
 [/INST] Using CS-inequality: let $x = a+b+c \implies \left(x+\frac{19}{2}\right)^2 = \left(1\sqrt{a+1}+2\sqrt{b+2}+3\sqrt{c+3}\right)^2\leq 14(x+6)\implies \left(x+\frac{5}{2}\right)^2 \leq 0\implies x = -\dfrac{5}{2}$
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