<s>[INST] If $x\geq 0,$ what is the smallest value of the function $f(x)= \frac{4x^2+ 8x + 13}{6(1+ x)}$ If $x\geq 0,$ what is the smallest value of the function
$$f(x)= \frac{4x^2+ 8x + 13}{6(1+ x)}$$
I tried doing it by completing the square in numerator and making it of the form
$$\frac{4(x+ 1)^2+ 9}{6(1+ x)}$$
and then, I put the value of $x= 0$ and the answer is coming out to be $13/6.$
But the actual answer is $2.$
Am I missing something ?
 [/INST] After completing the squares you can use the inequality between the arithmetic and geometric mean (AM-GM) in the form $a^2+b^2 \geq 2ab$:
Hence,
$$\frac{(2(x+1))^2+3^2}{6(x+1)}\stackrel{AM-GM}{\geq}\frac{2\cdot 2(x+1)\cdot 3}{6(x+1)}=2$$
Equality is reached for $2(x+1)=3 \Leftrightarrow x = \frac 12$.
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