Let $a=\cos\alpha$ and $b=\sin\alpha$, then \begin{align*} &a^4 + ab + b^4\\ &= \sin^4\alpha + \sin\alpha\cos\alpha + \cos^4\alpha\\ &= (\sin^2\alpha + \cos^2\alpha)^2 - 2\sin^2\alpha\cos^2\alpha + \sin\alpha\cos\alpha\\ &= 1 - \frac{1}{2}\sin^2\alpha + \frac{1}{2}\sin 2\alpha \end{align*} Let $u=\sin 2\alpha \in[-1, 1]$. Then the problem is equivalent to find the maximum value of the following quadratic polynomial when $-1 \le u\le 1$: $$f(u)=-\frac{1}{2} u ^2 +\frac{1}{2}u + 1$$ Because $u=\frac{\frac{1}{2}}{-2\cdot(-\frac{1}{2})}=\frac{1}{2}\in[-1,1]$, $f(u)$ reaches its maximum $\boxed{\frac{9}{8}}$ when $u=\frac{1}{2}$.