Question

If $\alpha, \beta, \gamma$ are the positive real roots of the equation $x^3-6 x^2+a x+b=0$ then the minimum value of $\frac{3}{\alpha+1}+\frac{3}{\beta+1}+\frac{3}{\gamma+1}$ is

• A. 1
• B. 3
• C. 9
• D. 27

Answer 1

Answer: (B)

$$ A.M. $\geq$ H.M.
$\Rightarrow \frac{\frac{3}{\alpha+1}+\frac{3}{\beta+1}+\frac{3}{\gamma+1}}{3} \geq \frac{3}{\frac{\alpha+1}{3}+\frac{\beta+1}{3}+\frac{\gamma+1}{3}}$
$\frac{3}{\alpha+1}+\frac{3}{\beta+1}+\frac{3}{\gamma+1} \geq \frac{3 \cdot 3 \cdot 3}{(\alpha+\beta+\gamma)+3} \geq \frac{27}{9} (\alpha+\beta+\gamma=6) $
$\Rightarrow \frac{3}{\alpha+1}+\frac{3}{\beta+1}+\frac{3}{\gamma+1} \geq 3$