Hard problem about law of the cosine

Question

I have been trying without success to prove by contradiction the following problem: Given 5 segments $x_1\leq x_2\leq x_3\leq x_4\leq x_5$ each three of which are sides of a triangle. Prove that there exists an acute angled triangle with sides chosen from the five given segments.

If we assume that all are obtuse angled triangles then we have:

$6x_5^2\geq 3x_3^2+3x_2^2+3x_1^2+3x_4^2$

$ 3x_4^2\geq 2x_3^2+2x_1^2+2x_2^2$

$x_3^2\geq x_2^2+x_1^2 $

However, I can't see the contradiction.

Answer 1

Suppose that there are no acute triangles. Then

• $x_1^2+x_2^2\le x_3^2$
• $x_2^2+x_3^2\le x_4^2$
• $x_3^2+x_4^2\le x_5^2$

But

$$(x_1+x_2)^2=x_1^2+x_2^2+2x_1x_2\le x_3^2+(x_1^2+x_2^2)\le x_3^2+x_4^2\le x_5^2$$

which contradicts the triangle inequality.

Answer 2

You are on the right track. By assuming that all the possible triples are associated with an obtuse triangle, it follows that:

$$ x_3^2 \geq x_1^2+x_2^2,\quad x_4^2\geq x_3^2+x_2^2 \geq x_1^2+2x_2^2,\quad x_5^2\geq x_4^2+x_3^2 \geq 2x_1^2+3x_2^2$$ but the last inequality contradicts: $$ x_5^2 \leq (x_1+x_2)^2 = x_1^2+2x_1 x_2+x_2^2 \leq x_1^2+3x_2^2.$$