For $x\geq 0$, what is the smallest value of $\frac{4x^2+8x+13}{6(x+1)}$?

Question

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!

Answer 1

$$\frac {4x^2+8x+13}{6(x+1)}$$

$$= \frac {2(x+1)}{3} + \frac {3}{2(x+1)}$$

$$ = y+\frac {1}{y} \ge 2$$

Answer 2

$$\frac{4x^2+8x+13}{6(x+1)} = \frac{4(x+1)^2+9}{6(x+1)} \ge \frac{2\sqrt{4(x+1)^2\cdot 9}}{6(x+1)} = \frac{2\cdot2\cdot3}{6} =\frac{12}{6} = 2$$

Answer 3

Solution

$$\frac{4x^2+8x+13}{6(x+1)}=\frac{4(x+1)^2+9}{6(x+1)}=\frac{2}{3}(x+1)+\frac{3}{2(x+1)}\geq 2$$

with the equality holding if and only if $\dfrac{2}{3}(x+1)=\dfrac{3}{2(x+1)}$,namely, $x=\dfrac{1}{2}.$

Answer 4

Hint: by polynomial euclidean division $\;\dfrac{4x^2+8x+13}{6(x+1)} = \dfrac{1}{6}\left(4(x+1) + \dfrac{9}{x+1}\right) \ge \ldots\,$