SKMO 2026/1 via KKT conditions
Problem (SKMO 2026, Problem 1). Let $x,y,z$ be positive real numbers such that
\[x^2+y^2+z^2=75.\]Suppose that two of the three sums $x+y$, $y+z$, $z+x$ are at least $10$. Determine the smallest and largest possible value of the remaining sum.
The natural toolkit is Lagrange multipliers with inequality constraints — the Karush–Kuhn–Tucker conditions. A purely algebraic argument also works (and we give one below), but KKT makes the structure transparent.
By symmetry we may assume
\[x+y\geq 10,\qquad y+z\geq 10,\]and ask for the smallest and largest possible value of $x+z$.
KKT conditions. For a minimization problem $\min f$ subject to an equality constraint $h=0$ and inequality constraints $g_i\leq 0$, a (regular) local minimum admits multipliers $\lambda\in\mathbb{R}$ and $\mu_i\geq 0$ such that
\[\nabla f + \lambda\,\nabla h + \sum_i \mu_i\,\nabla g_i = 0, \qquad \mu_i\, g_i = 0 \text{ for each } i.\]The slackness condition $\mu_i g_i = 0$ says that each inequality is either active ($g_i=0$) or has zero multiplier. The sign requirement $\mu_i\geq 0$ singles out minima; maxima satisfy the same equations with $\mu_i\leq 0$ (equivalently, replace $f$ by $-f$).
KKT formulation. We optimize $f(x,y,z)=x+z$ subject to the equality constraint $x^2+y^2+z^2=75$ and the inequality constraints
\[10-x-y\leq 0,\qquad 10-y-z\leq 0.\]For the minimization, the Lagrangian is
\[\mathcal L = x+z +\lambda(x^2+y^2+z^2-75) +\mu(10-x-y) +\nu(10-y-z), \qquad \mu,\nu\geq 0.\]The KKT stationarity conditions read
\[1+2\lambda x=\mu,\qquad 2\lambda y=\mu+\nu,\qquad 1+2\lambda z=\nu,\]together with the complementary slackness conditions $\mu(10-x-y)=0$ and $\nu(10-y-z)=0$.
Both inequality constraints are active. If both were inactive then $\mu=\nu=0$ and the middle equation forces $2\lambda y=0$; since $y>0$ this gives $\lambda=0$, but then the first equation becomes $1=0$. If only $x+y=10$ is active, then $\nu=0$, the third equation gives $\lambda=-\tfrac{1}{2z}<0$, and substituting into the middle equation yields $\mu=2\lambda y=-\tfrac{y}{z}<0$, contradicting $\mu\geq 0$. The other one-active case is symmetric.
Therefore $x+y=10$ and $y+z=10$, so $x=z$. Writing $x=z=a$ and $y=10-a$, the sphere condition becomes
\[2a^2 + (10-a)^2 = 75.\]Expanding,
\[2a^2 + 100 - 20a + a^2 = 75 \quad\Longleftrightarrow\quad 3a^2 - 20a + 25 = 0 \quad\Longleftrightarrow\quad (3a-5)(a-5)=0.\]So $a=5$ or $a=\tfrac{5}{3}$, giving two candidate triples
\[P_1 = (5,5,5), \qquad P_2 = \bigl(\tfrac{5}{3},\,\tfrac{25}{3},\,\tfrac{5}{3}\bigr),\]with $x+z=10$ at $P_1$ and $x+z=\tfrac{10}{3}$ at $P_2$.
Sign of the multipliers. Plugging each candidate into the KKT system,
\[P_1:\ \lambda=-\tfrac{1}{5},\ \mu=\nu=-1, \qquad P_2:\ \lambda=\tfrac{1}{5},\ \mu=\nu=\tfrac{5}{3}.\]Only $P_2$ has $\mu,\nu\geq 0$, so $P_2$ is the KKT point for the minimization. At $P_1$ the multipliers are reversed in sign — exactly the pattern required by the KKT system for the maximization (which has the right-hand sides of the stationarity conditions negated). Hence
\[\min(x+z) = \tfrac{10}{3}, \qquad \max(x+z) = 10,\]attained at $P_2$ and $P_1$ respectively.
A purely algebraic argument. KKT singles out the two candidates but, strictly speaking, only certifies that they are stationary; one still has to rule out the values outside $[\tfrac{10}{3},10]$. Here is a self-contained inequality proof. Write $s=x+z$. From $x\geq 10-y$ and $z\geq 10-y$,
\[\min(x,z)\geq 10-y,\]and since $\min(x,z)\leq \tfrac{s}{2}$,
\[y \geq 10-\tfrac{s}{2}.\]On the other hand $x^2+z^2\geq \tfrac{(x+z)^2}{2}=\tfrac{s^2}{2}$, so
\[y^2 \leq 75-\tfrac{s^2}{2}.\]Combining,
\[\left(10-\tfrac{s}{2}\right)^2 \leq 75-\tfrac{s^2}{2},\]which after expansion and multiplication by $4$ becomes
\[3s^2-40s+100 \leq 0, \quad\text{i.e.}\quad (3s-10)(s-10)\leq 0.\]Therefore $\tfrac{10}{3}\leq s\leq 10$, as KKT predicted.
The bounds are attained. The lower bound is attained at $x=z=\tfrac{5}{3}$, $y=\tfrac{25}{3}$: indeed
\[x^2+y^2+z^2 = \tfrac{25}{9}+\tfrac{625}{9}+\tfrac{25}{9} = \tfrac{675}{9} = 75,\]and $x+y = y+z = 10$, while $x+z = \tfrac{10}{3}$. The upper bound is attained at $x=y=z=5$, where all three pairwise sums equal $10$.
The smallest possible value of the remaining sum is $\tfrac{10}{3}$ and the largest is $10$.