Assume we have $6$ distinct points $A_1,A_2,...,A_6$ represents $6$ given irrational numbers $a_1,a_2,...,a_6$ such that no $3$ points are colinear. Coloring $\overline{A_iA_j}$ blue if $a_i+a_j$ is rational, red if $a_i+a_j$ is irrational. It's sufficient to prove there exists a triangle that has $3$ red sides. We will prove there is a triangle that its sides are the same color. Consider $A_1$. Since there are $5$ sides starting at $A_1$, by Pigenhole's theorem there exists $3$ sides that has the same color, assume they are blue. WLOG $A_1A_2,A_1A_3,A_1A_4$ are blue. Considering $\Delta A_2A_3A_4$. If its side are red, we have $\Delta A_2A_3A_4$ satisfied. If not, there exists a blue side. Assume $A_2A_3$ is blue, we have $\Delta A_1A_2A_3$ satisfied. Therefore, there is a triangle that its sides are the same color. Moreover, we can see that there is no triangle that has $3$ blue sides. Assume exists $\Delta A_iA_jA_k$. We have $a_i+a_j,a_j+a_k,a_k+a_i$ are rational, therefore $2a_j=(a_i+a_j)+(a_j+a_k)-(a_i+a_j)$ is rational, or $a_j$ is rational (contradiction) Hence, exists a triangle with $3$ red sides. (Q.E.D)