All are sides of obtuse triangles

What is the maximum number of positive integers among which any three are sides of an obtuse triangle?

I can find four, $11,11,16,20$. Is it possible to get five or more? We need $a^2+b^2<c^2$ and $a+b>c$ for all $a,b,c$.

Solution 1:

Suppose we have five positive integers $a_1\le a_2\le a_3\le a_4\le a_5$ such that any three are the sides of an obtuse triangle. Then by repeated substitution of $a_n^2+a_{n+1}^2<a_{n+2}^2$, $$a_5^2>a_4^2+a_3^2>2a_3^2+a_2^2>3a_2^2+2a_1^2$$ But $a_1+a_2>a_5$, so $$(a_1+a_2)^2>a_5^2>3a_2^2+2a_1^2=(a_2+a_1)^2+a_2^2+(a_2-a_1)^2$$which is impossible.