Every tetrahedron has a vertex whose adjacent edges can form a triangle 
Prove that in every tetrahedron there exists a vertex $v$ such that the three edges incident at $v$ have lengths that can form a triangle.

It can be proved using a tedious casework: assume by contradiction that all vertex have edges that do not satisfy the criterion, WLOG $a+b \leq c$ and there are $3$ cases for vertex B... It works out in the end, but is quite lengthy and boring.
Is there a nicer way? A friend suggested using geometric vectors and thus proving equalities instead of the triangle inequality, but I can't wrap it up.

 A: Lemma: Let $x,y,z>0$. Then there exists a triangle with sides $x,y,z$ if and only if $2x^2y^2+2y^2z^2+2z^2x^2>x^4+y^4+z^4$.
Proof of lemma: Observe that $2x^2y^2+2y^2z^2+2z^2x^2-x^4+y^4+z^4=(x+y+z)(-x+y+z)(x-y+z)(x+y-z)$. If there exists a triangle with sides $x,y,z$, then each factor is positive, so their product is positive. Conversely, if there is no triangle with sides $x,y,z$ then exactly one factor is nonpositive, so whole product is nonpositive.
Let's get back to the problem. Using notations from your drawing we have:
$$2a^2b^2+2b^2e^2+2e^2a^2>a^4+b^4+e^4, \\
2a^2c^2+2c^2d^2+2d^2a^2>a^4+c^4+d^4, \\
2b^2c^2+2c^2f^2+2f^2b^2>b^4+c^4+f^4, \\
2d^2e^2+2e^2f^2+2f^2d^2>d^4+e^4+f^4. $$
Summing up and dividing by 2 gives $$a^2b^2+b^2e^2+e^2a^2+a^2c^2+c^2d^2+d^2a^2+b^2c^2+c^2f^2+f^2b^2+d^2e^2+e^2f^2+f^2d^2 > a^4+b^4+c^4+d^4+e^4+f^4. \qquad (\spadesuit)$$
Assuming the contrary of thesis, we get 
$$2a^2b^2+2b^2c^2+2c^2a^2\le a^4+b^4+c^4, \\
2a^2d^2+2d^2e^2+2e^2a^2\le a^4+d^4+e^4, \\
2f^2c^2+2c^2d^2+2d^2f^2\le f^4+c^4+d^4, \\
2b^2e^2+2e^2f^2+2f^2b^2\le b^4+e^4+f^4. $$
Adding up and dividing by 2 yields 
$$a^2b^2+b^2c^2+c^2a^2+a^2d^2+d^2e^2+e^2a^2+f^2c^2+c^2d^2+d^2f^2+b^2e^2+e^2f^2+f^2b^2 \le a^4+b^4+c^4+d^4+e^4+f^4,$$
which contradicts $(\spadesuit)$!
