**Problem**: (TJ USAMO) Show that on any tetrahedron we can find a vertex $V$ such that the lengths of the $3$ edges with $V$ as an endpoint can be the sides of a triangle.

**Solution**: Call the tetrahedron $ ABCD $ and WLOG assume $ AB $ is the longest edge.

From triangle $ ABC $, we have $ BC+CA > AB $. From triangle $ ABD $, we have $ BD+DA > AB $. Summing the two inequalities together, we have

$ (BC+BD)+(CA+DA) > 2AB $.

It follows that one of $ BC+BD $ and $ CA+DA $ is greater than $ AB $. WLOG, assume $ CA+DA > AB $. But since $ AB $ is the longest edge and we have $ CA+DA > AB $, the edges $ AB $, $ AC $, and $ AD $ must form a triangle, as desired. QED.

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Comment: This problem was not too difficult; using the extremal principle to pick the longest side works well with triangles because as long as $ b+c > a $ when $ a $ is the largest of the three, $ a $, $ b $, and $ c $ can be the lengths of the sides of a triangle.

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Practice Problem: (TJ USAMO) Show that, for all positive reals $a$, $b$, and $c$ such that $a+b \ge c$; $b+c \ge a$; and $c+a \ge b$, we have

$2a^2(b + c) + 2b^2(c + a) + 2c^2(a + b) \ge a^3 + b^3 + c^3 + 9abc$.

When does equality hold?

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