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

Question

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 ?

Answer 1

The derivative, $$\frac{2}{3}-\frac{3}{2 (x+1)^2}$$ is zero when $x=1/2$:

Answer 2

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 $2(x+1)=3 \Leftrightarrow x = \frac 12$.

Answer 3

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 $2y-\dfrac3y=0\iff y^2=\dfrac32\iff x=?$