
4. Find the maximal value of $a^{2}+b^{2}+c^{2}$ for real numbers $a, b, c$ such that $a+b$, $b+c, c+a$ all lie in the interval $[0,1]$.

(Ján Mazák)


Solution. By symmetry of the problem, we can WLOG suppose $a \geqslant b \geqslant c$.

Then both $b-c$ and $b+c$ are non-negative (by our assumptions), hence their product will also be non-negative and $b^{2} \geqslant c^{2}$. Analogously, both $1-b-a$ and $1-b+a$ are non-negative and $(1-b)^{2} \geqslant a^{2}$.

Thus we have $a^{2}+b^{2}+c^{2} \leqslant(1-b)^{2}+2 b^{2}=1-b(2-3 b)$. The original restrictions on $a, b, c$ also impose certain restrictions on $b$ : we must have $1 \geqslant a+b \geqslant 2 b$ and $2 b \geqslant b+c \geqslant 0$, hence $b \in\left[0, \frac{1}{2}\right]$. But for $b$ in this interval it is clearly true that $b(2-3 b) \geqslant 0$, hence $a^{2}+b^{2}+c^{2} \leqslant 1$. This value is attained e.g. for $(a, b, c)=(1,0,0)$.

Remark. Note that we are maximizing a convex function on a convex set, therefore it is sufficient only to consider the boundary values where $a+b, b+c, c+a \in\{0,1\}$.

## Final Round of the 68th Czech and Slovak Mathematical Olympiad  (March 24-27, 2019)

![](https://cdn.mathpix.com/cropped/2024_04_17_447c393b5df10bbb25bcg-16.jpg?height=163&width=382&top_left_y=274&top_left_x=576)
