Question

The least value of $x^2+3y+1$ is

• A. 15
• B. Greater than 15
• C. 3
• D. Less than 15

Answer 1

Answer: (B)

Greater than 15

Given, $x^2{y}^3$ = 243 = $3^5$
Using $AM\ge HM$
$\frac{x^2+y+y+y+1}{5}\ge (x^2.y.y.y.1{)}^{1/5})$
$\Rightarrow x^2+3y+1\ge 5.\left(243{)}^{1/5}\right)\ge 15$
But $x^2,y\ne 1$, hence $x^2+3y+1>15$