jmc

The squared terms suggests the quadratic mean. Since we have no reciprocals or products, we can start with the QM-AM inequality on the numbers $a+b$, $b-c$, and $c-a$, which gives us $$\sqrt{\frac{(a+b{)}^2+(b-c{)}^2+(c-a{)}^2}{3}}\ge \frac{(a+b)+(b-c)+(c-a)}{3}=\frac{2b}{3}.$$Squaring both sides gives $$\frac{(a+b{)}^2+(b-c{)}^2+(c-a{)}^2}{3}\ge \frac{4b^2}{9}.$$Dividing both sides by $b^2$ and multiplting both sides by $3$ gives us $$\frac{(a+b{)}^2+(b-c{)}^2+(c-a{)}^2}{b^2}\ge \frac{4}{3}.$$Equality is achieved if $a+b=b-c=c-a$. From $a+b=b-c$ we get that $a=-c$. Then $a+b=c-a$ gives us $b=3c$. Hence if we pick $c=1$, $a=-1$, and $b=3$, we have $$\frac{(a+b{)}^2+(b-c{)}^2+(c-a{)}^2}{b^2}=\frac{(-1+3{)}^2+(3-1{)}^2+(1+1{)}^2}{3^2}=\frac{12}{9}=\boxed{\frac{4}{3}}.$$