I've found the following inequality $$\int_{B_r}\vert u\vert^q\leq C \bigg(\int_{B_r}\vert\nabla u\vert^2\bigg)^{a}\bigg(\int_{B_r}\vert u\vert ^2\bigg)^{\frac{q}{2}a}+\frac{c}{r^{2a}}\bigg(\int_{B_r}\vert u\vert ^2\bigg) ^{\frac{q}{2}}$$ for $u\in H^1(\mathbb{R}^3)$,$a=\frac{3}{4}(q2)$ and $q\in [2,6]$. Some hints to prove it? I've started using interpolation and Sobolev's inequality $$\int_{B_r}\vert u\vert^q\leq \Vert u\Vert_{L^2}^{q(1\theta)}\Vert u\Vert_{L^6}^{q \theta}\leq C \Vert u\Vert_{L^2}^{q(1\theta)}\Vert u\Vert_{H^1}^{q\theta}$$ with $\theta=\frac{3}{2}\frac{q2}{q}$. How can I go on?

Hint. For every $r>0$, and $q\le 6$, $$ W^{1,2}(B_r)\subset L^6(B_r)\subset L^q(B_r), $$ due to Sobolev Imbedding Theorem. In particular, there is a $c>0$, such that $$ \u\_{L^q(B_r)}^2 \le c_1 \big(\\nabla u\^2_{L^2(B_r)}+\u\^2_{L^2(B_r)}\big), $$ for all $u\in W^{1,2}(B_r)$. 

