( Inspired by USAMO 2014, P1 )

Let $P \in \mathbb{R}[X]$ be a monic polynomial of degree 2. Suppose that $P(1) \geq P(0)+3$, and that $P$ has two real roots (not necessarily distinct) $x_{1}$ and $x_{2}$. Find the smallest possible value for $\left(x_{1}^{2}+1\right)\left(x_{2}^{2}+1\right)$.
Let's write $P=X^{2}+a X+b$ with $a, b \in \mathbb{R}$. The condition in the statement becomes $1+a+b \geq b+3$ thus $a \geq 2$. Furthermore, by Vieta's formulas, $x_{1}+x_{2}=-a$ and $x_{1} x_{2}=b$. We write:

$$
\begin{aligned}
\left(x_{1}^{2}+1\right)\left(x_{2}^{2}+1\right) & =x_{1}^{2} x_{2}^{2}+x_{1}^{2}+x_{2}^{2}+1 \\
& =\left(x_{1} x_{2}\right)^{2}+\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}+1 \\
& =b^{2}+(-a)^{2}-2 b+1 \\
& =(b-1)^{2}+a^{2} \\
& \geq 0^{2}+2^{2} \\
& =4
\end{aligned}
$$

Thus, $\left(x_{1}^{2}+1\right)\left(x_{2}^{2}+1\right) \geq 4$. Moreover, we have equality for $a=2, b=1$, that is, for $P=$ $X^{2}+2 X+1=(X+1)^{2}$, which indeed satisfies $P(1)=4 \geq 4=3+P(0)$. The minimum value is therefore 4.

Alternative method: we apply the Cauchy-Schwarz inequality.

$$
\left(x_{1}^{2}+1\right)\left(x_{2}^{2}+1\right)=\left(x_{1}^{2}+1^{2}\right)\left(1^{2}+x_{2}^{2}\right) \underset{\substack{\text { Cauchy-Schwarz } \\ \text { inequality }}}{\geq}\left(x_{1} \times 1+1 \times x_{2}\right)^{2}=(-a)^{2}=a^{2} \geq 4
$$

Moreover, we have equality for $x_{1}=x_{2}=-1$, and we conclude similarly.