I am reading the appendix of Burq-Gérard-Tzvetkov: Multilinear eigenfunction estimates and global existence for the three dimensional NLS equations and there are some inequalities that they used and I am not able to prove them:
In the display (A.4) of page 53, why $$ \|g\|^{1/4}_{L^2} \|g\|^{3/4}_{H^2}\leq C n^{1/2}E_n(g)$$
where $E_n(g)=(n^2\|g\|^2_{L^2}+\frac{1}{n^2} \|\Delta g\|^2_{L^2})^{1/2}$. It seems to me like they are using Young's inequality, but I am not able to figure out how...
Second, between the display (A.6) and the display (A.7) why $$\|\nabla g \|_{H^{3/4}}\leq C n^{3/2}E_n(g) ?$$
Yes, the first inequality is indeed just the Young inequality for the product. If you write $a = \|g\|_{L^2}$ and $b = \|\Delta g\|_{L^2}$ then squaring everything, the inequality you want to prove can be written $$ a^{1/2}\,(a^2+b^2)^{3/4} \leq C\,n\,(n^2\,a^2 + b^2 / n^2). $$ Since $(a^2+b^2)^{3/4} \leq a^{3/2}+b^{3/2}$, it follows that $$ a^{1/2}\,(a^2+b^2)^{3/4} \leq a^{2} + a^{1/2}b^{3/2}. $$ Now $a^{2} \leq n\,n^2\,a^2$ because $n\geq 1$, while by Young's inequality $$ a^{1/2}b^{3/2} = n\,(na)^{1/2}\,(b/n)^{3/2} \leq n\left(\tfrac{n^2a^2}{4} + \tfrac{3\,b^2}{4\,n^2}\right), $$ hence $$ a^{1/2}\,(a^2+b^2)^{3/4} \leq \tfrac{5}{4}\,n\, \left(n^2a^2 + \tfrac{b^2}{n^2}\right). $$
Your second inequality follows similarly, noting indeed first that by interpolation $\|\nabla g\|_{H^{3/4}} \leq C\,(a + a^{1/4}\,b^{7/8})$ and so you get the result since $$ a^{1/8}\,b^{7/8} = n^{3/4} (na)^{1/8}\,(b/n)^{7/8}. $$ leads to $\|\nabla g\|_{H^{3/4}} \leq C\,n^{3/4} \,E_n(g)$.