jmc

Let $s=a+b.$ By QM-AM, $$\sqrt{\frac{a^2+b^2}{2}}\ge \frac{a+b}{2}=\frac{s}{2}.$$Then $\frac{a^2+b^2}{2}\ge \frac{s^2}{4},$ so $a^2+b^2\ge \frac{s^2}{2}.$ Hence, $$a^2+b^2+\frac{1}{(a+b{)}^2}\ge \frac{s^2}{2}+\frac{1}{s^2}.$$By AM-GM, $$\frac{s^2}{2}+\frac{1}{s^2}\ge 2\sqrt{\frac{s^2}{2}\cdot \frac{1}{s^2}}=\sqrt{2}.$$Equality occurs when $a=b$ and $s^2=\sqrt{2}.$ The numbers $a=b=2^{-3/4}$ satisfy these conditions.

Therefore, the minimum value is $\boxed{\sqrt{2}}.$