2. Find the minimum value of $x^{2}+4 y^{2}-2 x$, where $x$ and $y$ are real numbers that satisfy $2 x+8 y=3$.
Answer: $-\frac{19}{20}$
Solution: By the Cauchy-Schwarz Inequality,
$$
[2(x-1)+4(2 y)]^{2} \leq\left(2^{2}+4^{2}\right)\left[(x-1)^{2}+4 y^{2}\right] \text {. }
$$

Now, with $2(x-1)+4(2 y)=2 x+8 y-2=3-2=1$, we obtain
$$
\begin{aligned}
x^{2}+4 y^{2}-2 x & =(x-1)^{2}+4 y^{2}-1 \\
& \geq \frac{[2(x-1)+4(2 y)]^{2}}{2^{2}+4^{2}}-1 \\
& =\frac{1}{20}-1 \\
& =-\frac{19}{20}
\end{aligned}
$$

The minimum $-\frac{19}{20}$ is indeed obtained with $x=\frac{11}{10}$ and $y=\frac{1}{10}$.