Let $a,b \in L^p(\mathbb{T}^2)$ for all $p \in [1,+\infty]$. How can one show that : $$\|a^3+3ab^2+3a^2b \|_{L^2} \leqslant C (\|a\|_{L^6}^3 + \|a\|_{L^{\infty}}\|b^2\|_{L^2}) \; ? $$
By the triangular inequality and easy computations, one have $$\|a^3+3ab^2 \|_{L^2} \leqslant C (\|a\|_{L^6}^3 + \|a\|_{L^{\infty}}\|b^2\|_{L^2}), $$ so that it remains to prove $\|a^2b\|_{L^2} \leqslant C (\|a\|_{L^6}^3 + \|a\|_{L^{\infty}}\|b^2\|_{L^2})$. How is it possible to prove it ?
You can compute straight away $$\|a^2b\|_{L^2}^2 = \int a^3 ab^2 \leq \frac{1}{2} \int a^6 + a^2b^4 = C( \|a\|_{L^6}^6 + \|ab^2\|_{L^2}^2) \leq C( \|a\|_{L^6}^6 + \|a\|_{L^\infty}^2\|b^2\|_{L^2}^2). $$ Take the square root on both sides $$ \|ab^2\|_{L^2} \leq C \sqrt{\|a\|_{L^6}^6 + \|a\|_{L^\infty}^2\|b^2\|_{L^2}^2} \leq C(\|a\|_{L^6}^3 + \|a\|_{L^\infty}\|b^2\|_{L^2}). $$ Note that all we used where pointwise elementary inequality such as $$ ab \leq \frac{1}{2}(a^2+b^2) $$ and $$ \sqrt{a^2+b^2}\leq C (|a|+|b|). $$ For this reason this holds even on unbounded domains and not just on the torus.