## Task A-1.4.

Let $a, b$, and $c$ be real numbers that satisfy the equation

$$
|a+b|+|a+c|+|b+c|=8
$$

Determine the maximum and minimum possible value of the expression

$$
a^{2}+b^{2}+c^{2}
$$

and determine when it is achieved.
## Solution.

According to the triangle inequality, we have

$$
8=|a+b|+|b+c|+|c+a|=|a+b|+|a+c|+|-(b+c)| \geqslant|a+b+a+c-b-c|=|2 a|
$$

thus we conclude $|a| \leqslant 4$. Similarly, we get $|b| \leqslant 4$ and $|c| \leqslant 4$, hence

$$
a^{2}+b^{2}+c^{2} \leqslant 4^{2}+4^{2}+4^{2}=48
$$

One choice of $a, b$, and $c$ that achieves the value 48 is $a=4, b=c=-4$, so this is the maximum value of the expression.

For the minimum value, first apply the inequality between the quadratic and arithmetic means to the expressions $|a+b|,|b+c|$, and $|c+a|$:

$$
\sqrt{\frac{|a+b|^{2}+|b+c|^{2}+|c+a|^{2}}{3}} \geqslant \frac{|a+b|+|b+c|+|c+a|}{3}=\frac{8}{3}
$$

We can apply the inequality

$$
|x+y|^{2}=(x+y)^{2} \leqslant 2 x^{2}+2 y^{2}
$$

(which is equivalent to $(x-y)^{2} \geqslant 0$) to each of the terms in the numerator under the square root on the left side. This gives us

$$
\begin{aligned}
\sqrt{\frac{4\left(a^{2}+b^{2}+c^{2}\right)}{3}} & =\sqrt{\frac{2\left(\left(a^{2}+b^{2}\right)+\left(b^{2}+c^{2}\right)+\left(c^{2}+a^{2}\right)\right)}{3}} \\
& \geqslant \sqrt{\frac{(a+b)^{2}+(b+c)^{2}+(c+a)^{2}}{3}} \\
& =\sqrt{\frac{|a+b|^{2}+|b+c|^{2}+|c+a|^{2}}{3}} \geqslant \frac{8}{3}
\end{aligned}
$$

Rearranging, we get the inequality

$$
a^{2}+b^{2}+c^{2} \geqslant \frac{16}{3}
$$

Equality is achieved when all numbers $a, b$, and $c$ are equal to $\frac{4}{3}$ or $-\frac{4}{3}$, so the minimum value of the expression is indeed $\frac{16}{3}$.

Therefore, the minimum value of the given expression is $\frac{16}{3}$, and the maximum is 48.