Question

Sol. [Key Concept: Cubic with roots in H.P. ⇒ reciprocals in A.P.; relate to coefficients.]

$$\frac{1}{\alpha} + \frac{1}{\gamma} = \frac{2}{\beta} \Rightarrow \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{3}{\beta} = \frac{q}{q} = 1\ \Rightarrow \beta = 3$$

$$\frac{1}{\alpha^{2}} + \frac{1}{\beta^{2}} + \frac{1}{\gamma^{2}} \geq \frac{1}{\alpha\beta} + \frac{1}{\beta\gamma} + \frac{1}{\gamma\alpha}\ \Rightarrow 3\left( \frac{1}{\alpha^{2}} + \frac{1}{\beta^{2}} + \frac{1}{\gamma^{2}} \right) \geq \left( \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} \right)^{2} = 1$$

$$\Rightarrow \ \frac{1}{\alpha^{2}} + \frac{1}{\beta^{2}} + \frac{1}{\gamma^{2}} \geq \frac{1}{3} $$Again $\frac{1}{\alpha^{2}} + \frac{1}{\beta^{2}} + \frac{1}{\gamma^{2}} \geq \frac{1}{\alpha\beta} + \frac{1}{\beta\gamma} + \frac{1}{\gamma\alpha} \Rightarrow 3\left( \frac{1}{\alpha\beta} + \frac{1}{\beta\gamma} + \frac{1}{\gamma\alpha} \right) \leq \left( \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} \right)^{2} = 1$

Put $\beta = 3$ in the equation $27 + 9p + 3q - q = 0 \Rightarrow 9p + 2q + 27 = 0$

$$\frac{1}{\alpha\beta} + \frac{1}{\beta\gamma} + \frac{1}{\gamma\alpha} \leq \frac{1}{3} \Rightarrow \frac{\alpha + \beta + \gamma}{\alpha\beta\gamma} = \frac{- p}{q} \leq \frac{1}{3} \Rightarrow \frac{p}{q} \geq - \frac{1}{3}$$

✔ (a) $\frac{1}{\alpha^{2}} + \frac{1}{\beta^{2}} + \frac{1}{\gamma^{2}} \geq \frac{13}{}$

✔ (b) $9p + 2q + 27 = 0$

✔ (c) max of $\frac{1}{\alpha\beta} + \frac{1}{\beta\gamma} + \frac{1}{\gamma\alpha} = \frac{13}{}$

✔ (d) $pqge - \frac{13}{}$