10. (20 points) Find the smallest value of the parameter $c$ such that the system of equations has a unique solution

$$
\left\{\begin{array}{l}
8(x+7)^{4}+(y-4)^{4}=c \\
(x+4)^{4}+8(y-7)^{4}=c
\end{array}\right.
$$
Answer: $c=24$.

Solution. By the Cauchy-Bunyakovsky-Schwarz inequality, we have

(1) $\left(\frac{1}{2}+1\right)^{3}\left(8(x+\alpha)^{4}+(y-\beta)^{4}\right) \geqslant$

$$
\geqslant\left(\left(\frac{1}{2}+1\right)\left(2(x+\alpha)^{2}+(y-\beta)^{2}\right)\right)^{2} \geqslant
$$

$$
\geqslant(|x+\alpha|+|y-\beta|)^{4}
$$

(2) $\left(1+\frac{1}{2}\right)^{3}\left(8(x+\beta)^{4}+(y-\alpha)^{4}\right) \geqslant$

$$
\geqslant\left(\left(1+\frac{1}{2}\right)\left(2(x+\beta)^{2}+(y-\alpha)^{2}\right)\right)^{2} \geqslant
$$

$$
\geqslant(|x+\beta|+|y-\alpha|)^{4}
$$

Therefore, for any solution $(x, y)$ of the system, we have

(3) $\quad \frac{27}{4} c \geqslant(|x+\alpha|+|y-\beta|)^{4}+(|x+\beta|+|y-\alpha|)^{4} \geqslant$

$$
\begin{aligned}
& \geqslant(\alpha-\beta+x+y)^{4}+(\alpha-\beta-(x+y))^{4}= \\
& \quad=2(\alpha-\beta)^{4}+2(x+y)^{4}+12(\alpha-\beta)^{2}(x+y)^{2} \geqslant 2(\alpha-\beta)^{4}
\end{aligned}
$$

Thus, if $c<\frac{8}{27}(\alpha-\beta)^{4}$, the system has no solutions.

If $c=\frac{8}{27}(\alpha-\beta)^{4}$, then equality must be achieved in all inequalities (1)-(3). Therefore, $x+y=0$, the numbers $x+\alpha$ and $x+\beta$ must have different signs, and

$$
\frac{|x+\alpha|}{|x+\beta|}=\frac{1}{2}
$$

From these conditions, it follows that when $c=\frac{8}{27}(\alpha-\beta)^{4}$, there is only one solution $\left(-\frac{2 \alpha+\beta}{3}, \frac{2 \alpha+\beta}{3}\right)$. Remark. One can immediately notice that the solution of the system is symmetric with respect to the line $y=-x$ (two closed curves with centers at points $(-\alpha ; \beta)$ and $(-\beta ; \alpha)$). The solution will be unique if one of these curves touches this line. This is satisfied if $c=\min f(x)$, where $f(x)=8(x+\alpha)^{4}+(x+\beta)^{4} \cdot f^{\prime}(x)=32(x+\alpha)^{3}+4(x+\beta)^{3}=0$ at the point $x_{0}=-\frac{2 \alpha+\beta}{3}$. Then $c=f\left(x_{0}\right)$. https://ggbm.at/FVmUGS4Y