<s>[INST] For $x\geq 0$, what is the smallest value of $\frac{4x^2+8x+13}{6(x+1)}$? I know that I have to use the AM-GM inequality.
I tried separating the fraction: 
$$\frac{4x^2+2x+7}{6(x+1)} + \frac{6(x+1)}{6(x+1)}$$
However, it doesn't seem to make either side of the inequality into a number.
I would appreciate some help, thanks!
 [/INST] $$\frac{4x^2+8x+13}{6(x+1)} = \frac{4(x+1)^2+9}{6(x+1)} \ge \frac{2\sqrt{4(x+1)^2\cdot 9}}{6(x+1)} = \frac{2\cdot2\cdot3}{6} =\frac{12}{6} = 2$$
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