Does there exist $a,b,c\in \mathbb Q$ such that $(a+b+c)^2 + 3(a+b+c)+5=2(ab+bc+ca)$ 
Does there exist $ a,b,c\in \mathbb Q$ such that $(a+b+c)^2 + 3(a+b+c)+5=2(ab+bc+ca)$

I think the answer is no

After simplification of the equation you will get $$(a+\frac{3}{2})^2+(b+\frac{3}{2})^2+(c+\frac{3}{2})^2=\frac{7}{4}.$$Note that LHS$\geq 3.(\frac{3}{2})^2>RHS$, hence not possible.
