Question

For $x\ge 0$, the smallest value of expression $y=\frac{3x^2+6x+6}{2(x+1)}$ is

• A. 1
• B. 3
• C. 5
• D. 7

Answer 1

Answer: (B)

$y=\frac{3\left(x^2+2x+1\right)+3}{2(x+1)}$
$y=\frac{3(x+1{)}^2}{2(x+1)}+\frac{3}{2(x+1)}$
$y=\frac{3}{2}(x+1)+\frac{3}{2(x+1)}$
$y_{\min }\ge 2\sqrt{\frac{3}{2}\times \frac{3}{2}}$
$y_{\min }\ge 3$