
Problem 9.1. Let $p$ be a real parameter such that the equation $x^{2}-3 p x-p=0$ has real and distinct roots $x_{1}$ and $x_{2}$.

a) Prove that $3 p x_{1}+x_{2}^{2}-p>0$.

b) Find the least possible value of

$$
A=\frac{p^{2}}{3 p x_{1}+x_{2}^{2}+3 p}+\frac{3 p x_{2}+x_{1}^{2}+3 p}{p^{2}}
$$

When does equality obtain?


Solution: a) It follows from the equation that $x_{2}^{2}=3 p x_{2}+p$ and so $3 p x_{1}+x_{2}^{2}-p=3 p\left(x_{1}+x_{2}\right)=9 p^{2}>0$. The inequality is strict because otherwise $x_{1}=x_{2}=0$.

b) As in a), we obtain $3 p x_{1}+x_{2}^{2}+3 p=3 p x_{2}+x_{1}^{2}+3 p=$ $9 p^{2}+4 p>0$ (the last inequality follows from the conditions of the problem $x_{1}$ and $x_{2}$ to be distinct and real, giving $p \neq 0$ ). Therefore

$$
A=\frac{p^{2}}{9 p^{2}+4 p}+\frac{9 p^{2}+4 p}{p^{2}} \geq 2
$$

(from the Arithmetic-Geometric Mean Inequality) and equality obtains when $9 p^{2}+4 p=p^{2}$, i. e., when $p=-1 / 2$.
