Find the greatest value $M$ that the expression $7 x+10 y+z$ can take when $x, y, z$ are real numbers satisfying $x^{2}+2 x+\frac{1}{5} y^{2}+7 z^{2}=6$. In which cases is this value achieved? Go to the puzzle at the cell corresponding to $M$.

Go to the puzzle at the cell corresponding to $M$.
The constraint can be rewritten as $(x+1)^{2}+\frac{1}{5} y^{2}+7 z^{2}=7$. We will use Cauchy-Schwarz:

$$
\begin{aligned}
(7(x+1)+10 y+z)^{2} & \leq\left(7^{2}+5 \cdot 10^{2}+\frac{1}{7}\right)\left((x+1)^{2}+\frac{1}{5} y^{2}+7 z^{2}\right) \\
& =\left(549+\frac{1}{7}\right) \cdot 7=3844=62^{2}
\end{aligned}
$$

The maximum is achieved when $\left((x+1), \frac{1}{\sqrt{5}} y, \sqrt{7} z\right)$ is proportional to $\left(7,10 \sqrt{5}, \frac{1}{\sqrt{7}}\right)$, where the proportionality constant is chosen so that $(x, y, z)$ satisfies the imposed constraint. We then find $M=62-7=55$, with a unique case of equality: $(x, y, z)=(-13 / 62,175 / 31,1 / 62)$.