Let $$A=\left\{ (x_1,x_2, \dots, x_{10}) \in \mathbb R^{10}: \sum_{1 \le i < j \le 10} x_ix_j=45, x_1 \ge 0, x_2 \ge 0, \dots, x_{10} \ge 0 \right\}$$ and let $f(x_1,x_2, \dots, x_{10})=x_1+x_2+\dots + x_{10}$. Find infimum and supremum of the function $f$ on the set $A$; indicate which of these is achieved.
The most mechanical solution seems to be the use of the method of Lagrange multipliers. But it will be a monotonous calculation, and I feel that it can be done in a more brilliant way. Anyone got an idea?
The supremum is clearly $+\infty$. For example, let $x_j=0$ for all $j\ge 3$. Then the problem becomes maximizing $x_1+x_2$ subject to $x_1x_2=45$ and $x_1,x_2\ge 0$. There always exists some $x_2$ satisfying the constraints for any $x_1$ arbitrarily large, so the supremum is $+\infty$.
By using Maclaurin's inequalities, we have $$ \frac{\sum_{i=1}^{10} x_i}{10}\ge \left( \frac{\sum_{1\le i<j\le 10}x_ix_j}{45} \right)^{\frac{1}{2}}=1 $$ and the equality is attained if and only if $x_i=x_j$ for all $i,j$. The infimum is thus $10$, attained at $(x_1,\dots,x_{10})=(1,\dots,1)$.