Find the minimum.

Meaningless.

$\large{\sqrt[3]{11+\frac{1}{a^4}}+ \sqrt[3]{11+\frac{1}{b^4}} \geq 2\sqrt[6]{(11+\frac{1}{a^4})(11+\frac{1}{b^4})}}$

$= \sqrt[6]{121 + \dfrac{1}{a^4b^4} + \dfrac{11(a^4+b^4)}{a^4b^4}}$

Now, $a+b=1 \implies ab \leq \dfrac{1}{4} \implies \frac{1}{ab} \geq 4$

So, $\dfrac{1}{a^4b^4 }\geq 4^4 = 256$ and $a^4 + b^4 \geq 2a^2b^2$

$\implies$ $\dfrac{(11(a^4+b^4)}{a^4b^4} \geq \dfrac{11 \times 2}{a^2b^2} =$

$\dfrac{22}{a^2b^2} \geq 22 \times 4^2 = 22 \times 16 = 352$

Therefore $\large{2\sqrt[6]{121 + \dfrac{1}{a^4b^4} + \dfrac{11(a^4+b^4)}{a^4b^4}} \geq \sqrt[6]{121+264+352} = 2\sqrt[6]{729} = 2 \times 3 = 6}$

Hence $\large{\sqrt[3]{11+\frac{1}{a^4}}+ \sqrt[3]{11+\frac{1}{b^4}} \geq 6}$

Hence $\large{\sqrt[3]{11+\frac{1}{a^4}}+ \sqrt[3]{11+\frac{1}{b^4}} > 5}$

useless...

Absolutely makes no sense!