<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] Let $x+1=y^2$  as $x\ge0, y^2\ge1$
$$\dfrac{4x^2+8x+13}{1+x}=\dfrac{4(y^2-1)^2+8(y^2-1)+13}{y^2}=4y^2+\dfrac9{y^2}=\left(2y-\dfrac3y\right)^2+2\cdot2y\cdot\dfrac3y\ge12$$  the equality occurs if $2y-\dfrac3y=0\iff y^2=\dfrac32\iff x=?$
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