3. Given that $a, b$ are real numbers, satisfying
$$
t=\frac{a^{2}}{a^{2}+2 b^{2}}+\frac{b^{2}}{2 a^{2}+b^{2}} \text {. }
$$

Then the minimum value of $t$ is
3. $\frac{2}{3}$.

Let $x^{2}=a^{2}+2 b^{2}, y^{2}=2 a^{2}+b^{2}$. Then $a^{2}=\frac{-x^{2}+2 y^{2}}{3}, b^{2}=\frac{-y^{2}+2 x^{2}}{3}$.
Thus, $t=\frac{2 y^{2}-x^{2}}{3 x^{2}}+\frac{2 x^{2}-y^{2}}{3 y^{2}}$ $=\frac{2}{3}\left(\frac{y^{2}}{x^{2}}+\frac{x^{2}}{y^{2}}\right)-\frac{2}{3} \geqslant \frac{2}{3}$.
When and only when $\frac{y^{2}}{x^{2}}=\frac{x^{2}}{y^{2}}$, i.e., $a^{2}=b^{2}$, $t$ has the minimum value $\frac{2}{3}$.